5.17: The Beta Distribution
In this section, we will study the beta distribution, the most important distribution that has bounded support. But before we can study the beta distribution we must study the beta function .
The Beta Function
Definition
The beta function \( B \) is defined as follows: \[ B(a, b) = \int_0^1 u^{a-1} (1 - u)^{b - 1} du; \quad a, \, b \in (0, \infty) \]
Proof that \( B \) is well defined
We need to show that \(B(a, b) \lt \infty\) for every \(a, \, b \in (0, \infty)\). The integrand is positive on \( (0, 1) \), so the integral exists, either as a real number or \( \infty \). If \( a \ge 1 \) and \( b \ge 1 \), the integrand is continuous on \( [0, 1] \), so of course the integral is finite. Thus, the only cases of interest are when \( 0 \lt a \lt 1 \) or \( 0 \lt b \lt 1 \). Note that \[ \int_0^1 u^{a-1} (1 - u)^{b - 1} du = \int_0^{1/2} u^{a-1} (1 - u)^{b - 1} du + \int_{1/2}^1 u^{a-1} (1 - u)^{b - 1} du \] If \(0 \lt a \lt 1\), \((1 - u)^{b-1}\) is bounded on \(\left(0, \frac{1}{2}\right]\) and \( \int_0^{1/2} u^{a - 1} \, du = \frac{1}{a 2^a} \). Hence the first integral on the right in the displayed equation is finite. Similarly, If \(0 \lt b \lt 1\), \(u^{a-1}\) is bounded on \(\left[\frac{1}{2}, 1\right)\) and \( \int_{1/2}^1 (1 - u)^{b-1} \, du = \frac{1}{b 2^b} \). Hence the second integral on the right in the displayed equation is also finite.
The beta function was first introduced by Leonhard Euler.
Properties
The beta function satisfies the following properties:
- \(B(a, b) = B(b, a)\) for \( a, \, b \in (0, \infty) \), so \( B \) is symmetric.
- \(B(a, 1) = \frac{1}{a}\) for \( a \in (0, \infty) \)
- \( B(1, b) = \frac{1}{b} \) for \( b \in (0, \infty) \)
Proof
- Using the substitution \( v = 1 - u \) we have \[ B(a, b) = \int_0^1 u^{a-1}(1 - u)^{b-1} du = \int_0^1 (1 - v)^{a-1} v^{b-1} dv = B(b, a) \]
- \( B(a, 1) = \int_0^1 u^{a-1} du = \frac{1}{a} \)
- This follows from (a) and (b).
The beta function has a simple expression in terms of the gamma function:
If \( a, \, b \in (0, \infty) \) then \[ B(a, b) = \frac{\Gamma(a) \Gamma(b)}{\Gamma(a + b)} \]
Proof
From the definitions, we can express \(\Gamma(a + b) B(a, b)\) as a double integral: \[ \Gamma(a + b) B(a, b) = \int_0^\infty x^{a + b -1} e^{-x} dx \int_0^1 y^{a-1} (1 - y)^{b-1} dy = \int_0^\infty \int_0^1 (x y)^{a-1}[x (1 - y)]^{b-1} x e^{-x} dx \, dy \] Next we use the transformation \(w = x y\), \(z = x (1 - y)\) which maps \((0, \infty) \times (0, 1) \) one-to-one onto \( (0, \infty) \times (0, \infty) \). The inverse transformation is \( x = w + z \), \( y = w \big/(w + z) \) and the absolute value of the Jacobian is \[ \left|\det\frac{\partial(x, y)}{\partial(w, z)}\right| = \frac{1}{(w + z)} \] Thus, using the change of variables theorem for multiple integrals, the integral above becomes \[ \int_0^\infty \int_0^\infty w^{a-1} z^{b-1} (w + z) e^{-(w + z)} \frac{1}{w + z} dw \, dz \] which after simplifying is \( \Gamma(a) \Gamma(b) \).
Recall that the gamma function is a generalization of the factorial function. Here is the corresponding result for the beta function:
If \(j, \, k \in \N_+\) then \[ B(j, k) = \frac{(j - 1)! (k - 1)!}{(j + k - 1)!} \]
Proof
Recall that \( \Gamma(n) = (n - 1)! \) for \( n \in \N_+ \), so this result follows from the previous one.
Let's generalize this result. First, recall from our study of combinatorial structures that for \(a \in \R\) and \(j \in \N\), the ascending power of base \( a \) and order \( j \) is \[ a^{[j]} = a (a + 1) \cdots [a + (j - 1)] \]
If \(a, \, b \in (0, \infty)\), and \(j, \, k \in \N\), then \[ \frac{B(a + j, b + k)}{B(a, b)} = \frac{a^{[j]} b^{[k]}}{(a + b)^{[j + k]}} \]
Proof
Recall that \( \Gamma(a + j) = a^{[j]} \Gamma(a) \), so the result follows from the representation above for the beta function in terms of the gamma function.
\(B\left(\frac{1}{2}, \frac{1}{2}\right) = \pi\).
Proof
The Incomplete Beta Function
The integral that defines the beta function can be generalized by changing the interval of integration from \((0, 1)\) to \((0, x)\) where \(x \in [0, 1]\).
The incomplete beta function is defined as follows \[ B(x; a, b) = \int_0^x u^{a-1} (1 - u)^{b-1} du, \quad x \in (0, 1); \; a, \, b \in (0, \infty) \]
Of course, the ordinary (complete) beta function is \(B(a, b) = B(1; a, b)\) for \( a, \, b \in (0, \infty) \).
The Standard Beta Distribution
Distribution Functions
The beta distributions are a family of continuous distributions on the interval \( (0, 1) \).
The (standard) beta distribution with left parameter \( a \in (0, \infty) \) and right parameter \( b \in (0, \infty) \) has probability density function \( f \) given by \[ f(x) = \frac{1}{B(a, b)} x^{a-1} (1 - x)^{b-1}, \quad x \in (0, 1) \]
Of course, the beta function is simply the normalizing constant, so it's clear that \( f \) is a valid probability density function. If \( a \ge 1 \), \( f \) is defined at 0, and if \( b \ge 1 \), \( f \) is defined at 1. In these cases, it's customary to extend the domain of \( f \) to these endpoints. The beta distribution is useful for modeling random probabilities and proportions, particularly in the context of Bayesian analysis. The distribution has just two parameters and yet a rich variety of shapes (so in particular, both parameters are shape parameters ). Qualitatively, the first order properties of \( f \) depend on whether each parameter is less than, equal to, or greater than 1.
For \( a, \, b \in (0, \infty) \) with \( a + b \ne 2 \), define \[ x_0 = \frac{a - 1}{a + b - 2} \]
- If \(0 \lt a \lt 1\) and \(0 \lt b \lt 1\), \( f \) decreases and then increases with minimum value at \(x_0 \) and with \( f(x) \to \infty \) as \( x \downarrow 0 \) and as \( x \uparrow 1 \).
- If \(a = 1\) and \(b = 1\), \( f \) is constant.
- If \(0 \lt a \lt 1\) and \(b \ge 1\), \( f \) is decreasing with \( f(x) \to \infty \) as \( x \downarrow 0 \).
- If \(a \ge 1\) and \(0 \lt b \lt 1\), \( f \) is increasing with \( f(x) \to \infty \) as \( x \uparrow 1 \).
- If \(a = 1\) and \(b \gt 1\), \( f \) is decreasing with mode at \( x = 0 \).
- If \(a \gt 1\) and \(b = 1\), \( f \) is increasing with mode at \( x = 1 \).
- If \(a \gt 1\) and \(b \gt 1\), \( f \) increases and then decreases with mode at \( x_0 \).
Proof
These results follow from standard calculus. The first derivative is \[ f^\prime(x) = \frac{1}{B(a, b)} x^{a - 2}(1 - x)^{b - 2} [(a - 1) - (a + b - 2) x], \quad 0 \lt x \lt 1 \]
From part (b), note that the special case \(a = 1\) and \(b = 1\) gives the continuous uniform distribution on the interval \((0, 1)\) (the standard uniform distribution ). Note also that when \(a \lt 1\) or \(b \lt 1\), the probability density function is unbounded, and hence the distribution has no mode. On the other hand, if \(a \ge 1\), \(b \ge 1\), and one of the inequalites is strict, the distribution has a unique mode at \( x_0 \). The second order properties are more complicated.
For \( a, \, b \in (0, \infty) \) with \( a + b \notin \{2, 3\} \) and \( (a - 1)(b - 1)(a + b - 3) \ge 0 \), define \begin{align} x_1 &= \frac{(a - 1)(a + b - 3) - \sqrt{(a - 1)(b - 1)(a + b - 3)}}{(a + b - 3)(a + b - 2)}\\ x_2 &= \frac{(a - 1)(a + b - 3) + \sqrt{(a - 1)(b - 1)(a + b - 3)}}{(a + b - 3)(a + b - 2)} \end{align} For \( a \lt 1 \) and \( a + b = 2 \) or for \( b \lt 1 \) and \( a + b = 2 \), define \( x_1 = x_2 = 1 - a / 2 \).
- If \( a \le 1 \) and \( b \le 1 \), or if \( a \le 1 \) and \( b \ge 2 \), or if \( a \ge 2 \) and \( b \le 1 \), \( f \) is concave upward.
- If \( a \le 1 \) and \( 1 \lt b \lt 2 \), \( f \) is concave upward and then downward with inflection point at \( x_1 \).
- If \( 1 \lt a \lt 2 \) and \( b \le 1 \), \( f \) is concave downward and then upward with inflection point at \( x_2 \).
- If \( 1 \lt a \le 2 \) and \( 1 \lt b \le 2 \), \( f \) is concave downward.
- If \( 1 \lt a \le 2 \) and \( b \gt 2 \), \( f \) is concave downward and then upward with inflection point at \( x_2 \).
- If \( a \gt 2 \) and \( 1 \lt b \le 2 \), \( f \) is concave upward and then downward with inflection point at \( x_1 \).
- If \( a \gt 2 \) and \( b \gt 2 \), \( f \) is concave upward, then downward, then upward again, with inflection points at \( x_1 \) and \( x_2 \).
Proof
These results follow from standard (but very tedious) calculus. The second derivative is \[ f^{\prime\prime}(x) = \frac{1}{B(a, b)} x^{a - 3}(1 - x)^{b - 3} \left[(a + b - 2)(a + b - 3) x^2 - 2 (a - 1)(a + b - 3) x + (a - 1)(a - 2)\right] \]
In the special distribution simulator, select the beta distribution. Vary the parameters and note the shape of the beta density function. For selected values of the parameters, run the simulation 1000 times and compare the empirical density function to the true density function.
The special case \(a = \frac{1}{2}\), \(b = \frac{1}{2}\) is the arcsine distribution , with probability density function given by \[ f(x) = \frac{1}{\pi \, \sqrt{x (1 - x)}}, \quad x \in (0, 1) \] This distribution is important in a number of applications, and so the arcsine distribution is studied in a separate section.
The beta distribution function \(F\) can be easily expressed in terms of the incomplete beta function . As usual \(a\) denotes the left parameter and \(b\) the right parameter.
The beta distribution function \( F \) with parameters \(a, \, b \in (0, \infty)\) is given by \[ F(x) = \frac{B(x; a, b)}{B(a, b)}, \quad x \in (0, 1) \]
The distribution function \( F \) is sometimes known as the regularized incomplete beta function . In some special cases, the distribution function \(F\) and its inverse, the quantile function \(F^{-1}\), can be computed in closed form, without resorting to special functions.
If \(a \in (0, \infty)\) and \(b = 1\) then
- \(F(x) = x^a\) for \(x \in (0, 1)\)
- \(F^{-1}(p) = p^{1/a}\) for \(p \in (0, 1)\)
If \(a = 1\) and \(b \in (0, \infty)\) then
- \(F(x) = 1 - (1 - x)^b\) for \(x \in (0, 1)\)
- \(F^{-1}(p) = 1 - (1 - p)^{1/b}\) for \(p \in (0, 1)\)
If \(a = b = \frac{1}{2}\) (the arcsine distribution) then
- \(F(x) = \frac{2}{\pi} \arcsin\left(\sqrt{x}\right)\) for \(x \in (0, 1)\)
- \(F^{-1}(p) = \sin^2\left(\frac{\pi}{2} p\right)\) for \(p \in (0, 1)\)
There is an interesting relationship between the distribution functions of the beta distribution and the binomial distribution, when the beta parameters are positive integers. To state the relationship we need to embellish our notation to indicate the dependence on the parameters. Thus, let \(F_{a, b}\) denote the beta distribution function with left parameter \(a \in (0, \infty)\) and right parameter \(b \in (0, \infty)\), and let \(G_{n,p}\) denote the binomial distribution function with trial parameter \(n \in \N_+\) and success parameter \(p \in (0, 1)\).
If \(j, \, k \in \N_+\) and \(x \in (0, 1)\) then \[ F_{j,k}(x) = G_{j + k - 1, 1 - x}(k - 1) \]
Proof
By definition \[ F_{j,k}(x) = \frac{1}{B(j,k)} \int_0^x t^{j-1} (1 - t)^{k-1} dt \] Integrate by parts with \( u = (1 - t)^{k-1} \) and \( dv = t^{j-1} dt \), so that \( du = -(k -1) (1 - t)^{k-2} \) and \( v = t^j / j \). The result is \[ F_{j,k}(x) = \frac{1}{j B(j, k)} (1 - x)^{k-1} x^j + \frac{k-1}{j B(j, k)} \int_0^x t^j (1 - t)^{k-2} dt\] But by the property of the beta function above, \( B(j, k) = (j - 1)! (k - 1)! \big/(j + k - 1)! \). Hence \( 1 \big/ j B(j, k) = \binom{j + k - 1}{k - 1} \) and \( (k - 1) \big/ j B(j, k) = 1 \big/ B(j + 1, k - 1) \). Thus, the last displayed equation can be rewritten as \[F_{j,k}(x) = \binom{j + k - 1}{k - 1} (1 - x)^{k-1} x^j + F_{j + 1, k - 1}(x)\] Recall from the special case above that \(F_{j + k - 1, 1}(x) = x^{j + k - 1}\). Iterating the last displayed equation gives the result.
In the special distribution calculator, select the beta distribution. Vary the parameters and note the shape of the density function and the distribution function. In each of the following cases, find the median, the first and third quartiles, and the interquartile range. Sketch the boxplot.
- \(a = 1\), \(b = 1\)
- \(a = 1\), \(b = 3\)
- \(a = 3\), \(b = 1\)
- \(a = 2\), \(b = 4\)
- \(a = 4\), \(b = 2\)
- \(a = 4\), \(b = 4\)
Moments
The moments of the beta distribution are easy to express in terms of the beta function. As before, suppose that \(X\) has the beta distribution with left parameter \(a \in (0, \infty)\) and right parameter \(b \in (0, \infty)\).
If \(k \in [0, \infty)\) then \[ \E\left(X^k\right) = \frac{B(a + k, b)}{B(a, b)} \] In particular, if \( k \in \N \) then \[ \E\left(X^k\right) = \frac{a^{[k]}}{(a + b)^{[k]}} \]
Proof
Note that \[ \E\left(X^k\right) = \int_0^1 x^k \frac{1}{B(a, b)} x^{a-1} (1 - x)^{b - 1} dx = \frac{1}{B(a, b)} \int_0^1 x^{a + k - 1} (1 - x)^{b - 1} dx = \frac{B(a + k, b)}{B(a, b)}\] If \( k \in \N \), the formula simplifies by the property of the beta function above.
From the general formula for the moments, it's straightforward to compute the mean, variance, skewness, and kurtosis.
The mean and variance of \( X \) are \begin{align} \E(X) &= \frac{a}{a + b} \\ \var(X) &= \frac{a b}{(a + b)^2 (a + b + 1)} \end{align}
Proof
The formula for the mean and variance follow from the formula for the moments and the computational formula \( \var(X) = \E(X^2) - [\E(X)]^2 \)
Note that the variance depends on the parameters \( a \) and \( b \) only through the product \( a b \) and the sum \( a + b \).
Open the special distribution simulator and select the beta distribution. Vary the parameters and note the size and location of the mean\(\pm\)standard deviation bar. For selected values of the parameters, run the simulation 1000 times and compare the sample mean and standard deviation to the distribution mean and standard deviation.
The skewness and kurtosis of \( X \) are \begin{align} \skw(X) &= \frac{2 (b - a) \sqrt{a + b + 1}}{(a + b + 2) \sqrt{a b}} \\ \kur(X) &= \frac{3 a^3 b + 3 a b^3 + 6 a^2 b^2 + a^3 + b^3 + 13 a^2 b + 13 a b^2 + a^2 + b^2 + 14 a b}{a b (a + b + 2) (a + b + 3)} \end{align}
Proof
These results follow from the computational formulas that give the skewness and kurtosis in terms of \( \E(X^k) \) for \( k \in \{1, 2, 3, 4\} \), and the formula for the moments above.
In particular, note that the distribution is positively skewed if \( a \lt b \), unskewed if \( a = b \) (the distribution is symmetric about \( x = \frac{1}{2} \) in this case) and negatively skewed if \( a \gt b \).
Open the special distribution simulator and select the beta distribution. Vary the parameters and note the shape of the probability density function in light of the previous result on skewness. For various values of the parameters, run the simulation 1000 times and compare the empirical density function to the true probability density function.
Related Distributions
The beta distribution is related to a number of other special distributions.
If \(X\) has the beta distribution with left parameter \(a \in (0, \infty)\) and right parameter \(b \in (0, \infty)\) then \(Y = 1 - X\) has the beta distribution with left parameter \(b\) and right parameter \(a\).
Proof
This follows from the standard change of variables formula. If \( f \) and \( g \) denote the PDFs of \( X \) and \( Y \) respectively, then \[ g(y) = f(1 - y) = \frac{1}{B(a, b)} (1 - y)^{a - 1} y^{b - 1} = \frac{1}{B(b, a)} y^{b - 1} (1 - y)^{a - 1}, \quad y \in (0, 1)\]
The beta distribution with right parameter 1 has a reciprocal relationship with the Pareto distribution.
Suppose that \( a \in (0, \infty) \).
- If \(X\) has the beta distribution with left parameter \(a \) and right parameter 1 then \(Y = 1 / X\) has the Pareto distribution with shape parameter \(a\).
- If \( Y \) has the Pareto distribution with shape parameter \( a \) then \( X = 1 / Y \) has the beta distribution with left parameter \( a \) and right parameter 1.
Proof
The two results are equivalent. In (a), suppose that \( X \) has the beta distribution with parameters \( a \) and 1. The transformation \( y = 1 / x \) maps \( (0, 1) \) one-to-one onto \( (0, \infty) \). The inverse is \( x = 1 / y \) with \( dx/dy = -1/y^2 \). Recall also that \( B(a, 1) = 1 / a \). By the change of variables formula, the PDF \( g \) of \( Y = 1 / X \) is given by \[ g(y) = f\left(\frac{1}{y}\right) \frac{1}{y^2} = a \left(\frac{1}{y}\right)^{a-1} \frac{1}{y^2} = \frac{a}{y^{a+1}}, \quad y \in (0, \infty) \] We recognize \( g \) as the PDF of the Pareto distribution with shape parameter \( a \).
The following result gives a connection between the beta distribution and the gamma distribution.
Suppose that \(X\) has the gamma distribution with shape parameter \(a \in (0, \infty)\) and rate parameter \(r \in (0, \infty)\), \(Y\) has the gamma distribution with shape parameter \(b \in (0, \infty)\) and rate parameter \(r\), and that \(X\) and \(Y\) are independent. Then \(V = X \big/ (X + Y)\) has the beta distribution with left parameter \(a\) and right parameter \(b\).
 We will actually prove stronger results: \( U \) and \( V \) are independent, \( U \) has the gamma distribution with shape parameter \( a + b \) and rate parameter \( r \), and \( V \) has the beta distribution with parameters \( a \) and \( b \). First note that \( (X, Y) \) has joint PDF \( f \) given by \[ f(x, y) = \frac{r^a}{\Gamma(a)} x^{a-1} e^{-r x} \frac{r^b}{\Gamma(b)} y^{b-1} e^{-r y} = \frac{r^{a+b}}{\Gamma(a) \Gamma(b)} x^{a-1} y^{b-1} e^{-r(x + y)}; \quad x, \, y \in (0, \infty) \] The transformation \( u = x + y \) and \( v = x \big/ (x + y) \) maps \( (0, \infty) \times (0, \infty) \) one-to-one onto \( (0, \infty) \times (0, 1) \). The inverse is \( x = u v \), \( y = u(1 - v) \) and the absolute value of the Jacobian is \[ \left|\det \frac{\partial(x, y)}{\partial(u, v)}\right| = u \] Hence by the multivariate change of variables theorem, the PDF \( g \) of \( (U, V) \) is given by \begin{align} g(u, v) & = f[u v, u(1 - v)] u = \frac{r^{a+b}}{\Gamma(a) \Gamma(b)} (u v)^{a-1} [u(1 - v)]^{b-1} e^{-ru} u \\ & = \frac{r^{a+b}}{\Gamma(a) \Gamma(b)} u^{a+b-1} e^{-ru} v^{a-1} (1 - v)^{b-1} \\ & = \frac{r^{a+b}}{\Gamma(a + b)} u^{a+b-1} e^{-ru} \frac{\Gamma(a + b)}{\Gamma(a) \Gamma(b)} v^{a-1} (1 - v)^{b-1}; \quad u \in (0, \infty), v \in (0, 1) \end{align} The results now follow from the factorization theorem. The factor in \( u \) is the gamma PDF with shape parameter \( a + b \) and rate parameter \( r \) while the factor in \( v \) is the beta PDF with parameters \( a \) and \( b \).
The following result gives a connection between the beta distribution and the \(F\) distribution. This connection is a minor variation of the previous result.
If \(X\) has the \(F\) distribution with \(n \in (0, \infty)\) degrees of freedom in the numerator and \(d \in (0, \infty)\) degrees of freedom in the denominator then \[ Y = \frac{(n / d) X}{1 + (n / d)X} \] has the beta distribution with left parameter \(a = n / 2\) and right parameter \(b = d / 2\).
Proof
If \(X\) has the \(F\) distribution with \(n \gt 0\) degrees of freedom in the numerator and \(d \gt 0\) degrees of freedom in the denominator then \( X \) can be written as \[ X = \frac{U / n}{V / d} \] where \( U \) has the chi-square distribution with \( n \) degrees of freedom, \( V \) has the chi-square distribution with \( d \) degrees of freedom, and \( U \) and \( V \) are independent. Hence \[ Y = \frac{(n / d) X}{1 + (n / d) X} = \frac{U / V}{1 + U / V} = \frac{U}{U + V} \] But the chi-square distribution is a special case of the gamma distribution. Specifically, \( U \) has the gamma distribution with shape parameter \( n / 2 \) and rate parameter \( 1/2 \), \( V \) has the gamma distribution with shape parameter \( d / 2 \) and rate parameter \( 1/2 \), and again \( U \) and \( V \) are independent. Hence by the previous result , \( Y \) has the beta distribution with left parameter \( n/2 \) and right parameter \( d/2 \).
Our next result is that the beta distribution is a member of the general exponential family of distributions.
Suppose that \(X\) has the beta distribution with left parameter \(a \in (0, \infty)\) and right parameter \(b \in (0, \infty)\). Then the distribution is a two-parameter exponential family with natural parameters \(a - 1\) and \(b - 1\), and natural statistics \(\ln(X)\) and \(\ln(1 - X)\).
Proof
This follows from the definition of the general exponential distribution, since the PDF \( f \) of \( X \) can be written as \[ f(x) = \frac{1}{B(a, b)} \exp\left[(a - 1) \ln(x) + (b - 1) \ln(1 - x)\right], \quad x \in (0, 1) \]
The beta distribution is also the distribution of the order statistics of a random sample from the standard uniform distribution.
Suppose \( n \in \N_+ \) and that \( (X_1, X_2, \ldots, X_n) \) is a sequence of independent variables, each with the standard uniform distribution. For \( k \in \{1, 2, \ldots, n\} \), the \( k \)th order statistics \( X_{(k)} \) has the beta distribution with left parameter \( a = k \) and right parameter \( b = n - k + 1 \).
Proof
See the section on order statistics.
One of the most important properties of the beta distribution, and one of the main reasons for its wide use in statistics, is that it forms a conjugate family for the success probability in the binomial and negative binomial distributions.
Suppose that \( P \) is a random probability having the beta distribution with left parameter \( a \in (0, \infty) \) and right parameter \( b \in (0, \infty) \). Suppose also that \( X \) is a random variable such that the conditional distribution of \( X \) given \( P = p \in (0, 1) \) is binomial with trial parameter \( n \in \N_+ \) and success parameter \( p \). Then the conditional distribution of \( P \) given \( X = k \) is beta with left parameter \( a + k \) and right parameter \( b + n - k \).
Proof
The joint PDF \( f \) of \( (P, X) \) on \( (0, 1) \times \{0, 1, \ldots n\} \) is given by \[ f(p, k) = \frac{1}{B(a, b)} p^{a-1} (1 - p)^{b-1} \binom{n}{k} p^k (1 - p)^{n-k} = \frac{1}{B(a, b)} \binom{n}{k} p^{a + k - 1} (1 - p)^{b + n - k - 1} \] The conditional PDF of \( P \) given \( X = k \) is simply the normalized version of the function \( p \mapsto f(p, k) \). We can tell from the functional form that this distribution is beta with the parameters given in the theorem.
Suppose again that \( P \) is a random probability having the beta distribution with left parameter \( a \in (0, \infty) \) and right parameter \( b \in (0, \infty) \). Suppose also that \( N \) is a random variable such that the conditional distribution of \( N \) given \( P = p \in (0, 1) \) is negative binomial with stopping parameter \( k \in \N_+ \) and success parameter \( p \). Then the conditional distribution of \( P \) given \( N = n \) is beta with left parameter \( a + k \) and right parameter \( b + n - k \).
Proof
The joint PDF \( f \) of \( (P, N) \) on \( (0, 1) \times \{k, k + 1, \ldots\} \) is given by \[ f(p, n) = \frac{1}{B(a, b)} p^{a-1} (1 - p)^{b-1} \binom{n - 1}{k - 1} p^k (1 - p)^{n-k} = \frac{1}{B(a, b)} \binom{n - 1}{k - 1} p^{a + k - 1} (1 - p)^{b + n - k - 1} \] The conditional PDF of \( P \) given \( N = n \) is simply the normalized version of the function \( p \mapsto f(p, n) \). We can tell from the functional form that this distribution is beta with the parameters given in the theorem.
in both cases, note that in the posterior distribution of \( P \), the left parameter is increased by the number of successes and the right parameter by the number of failures. For more on this, see the section on Bayesian estimation in the chapter on point estimation.
The General Beta Distribution
The beta distribution can be easily generalized from the support interval \((0, 1)\) to an arbitrary bounded interval using a linear transformation. Thus, this generalization is simply the location-scale family associated with the standard beta distribution.
Suppose that \(Z\) has the standard beta distibution with left parameter \(a \in (0, \infty)\) and right parameter \(b \in (0, \infty)\). For \(c \in \R\) and \(d \in (0, \infty)\) random variable \(X = c + d Z\) has the beta distribution with left parameter \( a \), right parameter \( b \), location parameter \( c \) and scale parameter \( d \).
For the remainder of this discussion, suppose that \( X \) has the distribution in the definition above.
\(X\) has probability density function
\[ f(x) = \frac{1}{B(a, b) d^{a + b - 1}} (x - c)^{a - 1} (c + d - x)^{b - 1}, \quad x \in (c, c + d) \]Proof
This follows from a standard result for location-scale families. If \( g \) denotes the standard beta PDF of \( Z \), then \( X \) has PDF \( f \) given by \[ f(x) = \frac{1}{d} g\left(\frac{x - c}{d}\right), \quad x \in (c, c + d) \]
Most of the results in the previous sections have simple extensions to the general beta distribution.
The mean and variance of \( X \) are
- \(\E(X) = c + d \frac{a}{a + b}\)
- \(\var(X) = d^2 \frac{a b}{(a + b)^2 (a + b + 1)}\)
Proof
This follows from the standard mean and variance and basic properties of expected value and variance.
- \( \E(X) = c + d \E(Z)\)
- \( \var(X) = d^2 \var(Z) \)
Recall that skewness and variance are defined in terms of standard scores, and hence are unchanged under location-scale transformations. Hence the skewness and kurtosis of \( X \) are just as for the standard beta distribution .


13.5E: Counting Principles (Exercises)
31. How many ways are there to choose a number from the set \{-10,-6,4,10,12,18,24,32\} that is divisible by either 4 or 6?
32. In a group of 20 musicians, 12 play piano, 7 play trumpet, and 2 play both piano and trumpet. How many musicians play either piano or trumpet?
33. How many ways are there to construct a 4 -digit code if numbers can be repeated?
34. A palette of water color paints has 3 shades of green, 3 shades of blue, 2 shades of red, 2 shades of yellow, and 1 shade of black. How many ways are there to choose one shade of each color?
35. Calculate \(P(18,4)\).
36. In a group of 5 freshman, 10 sophomores, 3 juniors, and 2 seniors, how many ways can a president, vice president, and treasurer be elected?
37. Calculate \(C(15,6)\).
38. A coffee shop has 7 Guatemalan roasts, 4 Cuban roasts, and 10 Costa Rican roasts. How many ways can the shop choose 2 Guatemalan, 2 Cuban, and 3 Costa Rican roasts for a coffee tasting event?
39. How many subsets does the set \(\{1,3,5, \ldots, 99\}\) have?
40. A day spa charges a basic day rate that includes use of a sauna, pool, and showers. For an extra charge, guests can choose from the following additional services: massage, body scrub, manicure, pedicure, facial, and straight-razor shave. How many ways are there to order additional services at the day spa?
41. How many distinct ways can the word DEADWOOD be arranged?
42. How many distinct rearrangements of the letters of the word DEADWOOD are there if the arrangement must begin and end with the letter D?


2.10: Descriptive Statistics (Exercises)
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2.2: Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
Q 2.2.1
Student grades on a chemistry exam were: 77, 78, 76, 81, 86, 51, 79, 82, 84, 99
- Construct a stem-and-leaf plot of the data.
- Are there any potential outliers? If so, which scores are they? Why do you consider them outliers?
Q 2.2.2
Table contains the 2010 obesity rates in U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
- Use a random number generator to randomly pick eight states. Construct a bar graph of the obesity rates of those eight states.
- Construct a bar graph for all the states beginning with the letter "A."
- Construct a bar graph for all the states beginning with the letter "M."
S 2.2.2
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Example solution for using the random number generator to generate a simple random sample of 8 states. Instructions are as follows.
Eight numbers are generated (use the right arrow key to scroll through the numbers). The numbers correspond to the numbered states (for this example: {47 21 9 23 51 13 25 4}. If any numbers are repeated, generate a different number by using 5:randInt(51,1)). Here, the states (and Washington DC) are {Arkansas, Washington DC, Idaho, Maryland, Michigan, Mississippi, Virginia, Wyoming}.
Corresponding percents are {30.1, 22.2, 26.5, 27.1, 30.9, 34.0, 26.0, 25.1}.
- Number the entries in the table 1–51 (Includes Washington, DC; Numbered vertically)
Figure \(\PageIndex{1}\): (a)
Figure \(\PageIndex{1}\): (b)
Figure \(\PageIndex{1}\): (c)
For each of the following data sets, create a stem plot and identify any outliers.
Exercise 2.2.7
The miles per gallon rating for 30 cars are shown below (lowest to highest).
19, 19, 19, 20, 21, 21, 25, 25, 25, 26, 26, 28, 29, 31, 31, 32, 32, 33, 34, 35, 36, 37, 37, 38, 38, 38, 38, 41, 43, 43
Answer
| Stem | Leaf |
|---|---|
| 1 | 9 9 9 |
| 2 | 0 1 1 5 5 5 6 6 8 9 |
| 3 | 1 1 2 2 3 4 5 6 7 7 8 8 8 8 |
| 4 | 1 3 3 |
The height in feet of 25 trees is shown below (lowest to highest).
25, 27, 33, 34, 34, 34, 35, 37, 37, 38, 39, 39, 39, 40, 41, 45, 46, 47, 49, 50, 50, 53, 53, 54, 54
The data are the prices of different laptops at an electronics store. Round each value to the nearest ten.
249, 249, 260, 265, 265, 280, 299, 299, 309, 319, 325, 326, 350, 350, 350, 365, 369, 389, 409, 459, 489, 559, 569, 570, 610
Answer
| Stem | Leaf |
|---|---|
| 2 | 5 5 6 7 7 8 |
| 3 | 0 0 1 2 3 3 5 5 5 7 7 9 |
| 4 | 1 6 9 |
| 5 | 6 7 7 |
| 6 | 1 |
The data are daily high temperatures in a town for one month.
61, 61, 62, 64, 66, 67, 67, 67, 68, 69, 70, 70, 70, 71, 71, 72, 74, 74, 74, 75, 75, 75, 76, 76, 77, 78, 78, 79, 79, 95
For the next three exercises, use the data to construct a line graph.
Exercise 2.2.8
In a survey, 40 people were asked how many times they visited a store before making a major purchase. The results are shown in Table .
| Number of times in store | Frequency |
|---|---|
| 1 | 4 |
| 2 | 10 |
| 3 | 16 |
| 4 | 6 |
| 5 | 4 |
Answer
Figure \(\PageIndex{3}\):
Exercise 2.2.9
In a survey, several people were asked how many years it has been since they purchased a mattress. The results are shown in Table .
| Years since last purchase | Frequency |
|---|---|
| 0 | 2 |
| 1 | 8 |
| 2 | 13 |
| 3 | 22 |
| 4 | 16 |
| 5 | 9 |
Exercise 2.2.10
Several children were asked how many TV shows they watch each day. The results of the survey are shown in Table .
| Number of TV Shows | Frequency |
|---|---|
| 0 | 12 |
| 1 | 18 |
| 2 | 36 |
| 3 | 7 |
| 4 | 2 |
Answer
Figure \(\PageIndex{3}\):
Exercise 2.2.11
The students in Ms. Ramirez’s math class have birthdays in each of the four seasons. Table shows the four seasons, the number of students who have birthdays in each season, and the percentage (%) of students in each group. Construct a bar graph showing the number of students.
| Seasons | Number of students | Proportion of population |
|---|---|---|
| Spring | 8 | 24% |
| Summer | 9 | 26% |
| Autumn | 11 | 32% |
| Winter | 6 | 18% |
Using the data from Mrs. Ramirez’s math class supplied in Exercise , construct a bar graph showing the percentages.
Answer
Figure \(\PageIndex{5}\):
Exercise 2.2.12
David County has six high schools. Each school sent students to participate in a county-wide science competition. Table shows the percentage breakdown of competitors from each school, and the percentage of the entire student population of the county that goes to each school. Construct a bar graph that shows the population percentage of competitors from each school.
| High School | Science competition population | Overall student population |
|---|---|---|
| Alabaster | 28.9% | 8.6% |
| Concordia | 7.6% | 23.2% |
| Genoa | 12.1% | 15.0% |
| Mocksville | 18.5% | 14.3% |
| Tynneson | 24.2% | 10.1% |
| West End | 8.7% | 28.8% |
Use the data from the David County science competition supplied in Exercise . Construct a bar graph that shows the county-wide population percentage of students at each school.
Answer
Figure <IP_ADDRESS>: Histograms, Frequency, Polygons, and Time Series Graphs
Q 2.3.1
Suppose that three book publishers were interested in the number of fiction paperbacks adult consumers purchase per month. Each publisher conducted a survey. In the survey, adult consumers were asked the number of fiction paperbacks they had purchased the previous month. The results are as follows:
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 10 | |
| 1 | 12 | |
| 2 | 16 | |
| 3 | 12 | |
| 4 | 8 | |
| 5 | 6 | |
| 6 | 2 | |
| 8 | 2 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0 | 18 | |
| 1 | 24 | |
| 2 | 24 | |
| 3 | 22 | |
| 4 | 15 | |
| 5 | 10 | |
| 7 | 5 | |
| 9 | 1 |
| # of books | Freq. | Rel. Freq. |
|---|---|---|
| 0–1 | 20 | |
| 2–3 | 35 | |
| 4–5 | 12 | |
| 6–7 | 2 | |
| 8–9 | 1 |
- Find the relative frequencies for each survey. Write them in the charts.
- Using either a graphing calculator, computer, or by hand, use the frequency column to construct a histogram for each publisher's survey. For Publishers A and B, make bar widths of one. For Publisher C, make bar widths of two.
- In complete sentences, give two reasons why the graphs for Publishers A and B are not identical.
- Would you have expected the graph for Publisher C to look like the other two graphs? Why or why not?
- Make new histograms for Publisher A and Publisher B. This time, make bar widths of two.
- Now, compare the graph for Publisher C to the new graphs for Publishers A and B. Are the graphs more similar or more different? Explain your answer.
Q 2.3.2
Often, cruise ships conduct all on-board transactions, with the exception of gambling, on a cashless basis. At the end of the cruise, guests pay one bill that covers all onboard transactions. Suppose that 60 single travelers and 70 couples were surveyed as to their on-board bills for a seven-day cruise from Los Angeles to the Mexican Riviera. Following is a summary of the bills for each group.
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 51–100 | 5 | |
| 101–150 | 10 | |
| 151–200 | 15 | |
| 201–250 | 15 | |
| 251–300 | 10 | |
| 301–350 | 5 |
| Amount($) | Frequency | Rel. Frequency |
|---|---|---|
| 100–150 | 5 | |
| 201–250 | 5 | |
| 251–300 | 5 | |
| 301–350 | 5 | |
| 351–400 | 10 | |
| 401–450 | 10 | |
| 451–500 | 10 | |
| 501–550 | 10 | |
| 551–600 | 5 | |
| 601–650 | 5 |
- Fill in the relative frequency for each group.
- Construct a histogram for the singles group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis.
- Construct a histogram for the couples group. Scale the x -axis by $50 widths. Use relative frequency on the y -axis.
-
Compare the two graphs:
- List two similarities between the graphs.
- List two differences between the graphs.
- Overall, are the graphs more similar or different?
- Construct a new graph for the couples by hand. Since each couple is paying for two individuals, instead of scaling the x -axis by $50, scale it by $100. Use relative frequency on the y -axis.
-
Compare the graph for the singles with the new graph for the couples:
- List two similarities between the graphs.
- Overall, are the graphs more similar or different?
- How did scaling the couples graph differently change the way you compared it to the singles graph?
- Based on the graphs, do you think that individuals spend the same amount, more or less, as singles as they do person by person as a couple? Explain why in one or two complete sentences.
S 2.3.2
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 51–100 | 5 | 0.08 |
| 101–150 | 10 | 0.17 |
| 151–200 | 15 | 0.25 |
| 201–250 | 15 | 0.25 |
| 251–300 | 10 | 0.17 |
| 301–350 | 5 | 0.08 |
| Amount($) | Frequency | Relative Frequency |
|---|---|---|
| 100–150 | 5 | 0.07 |
| 201–250 | 5 | 0.07 |
| 251–300 | 5 | 0.07 |
| 301–350 | 5 | 0.07 |
| 351–400 | 10 | 0.14 |
| 401–450 | 10 | 0.14 |
| 451–500 | 10 | 0.14 |
| 501–550 | 10 | 0.14 |
| 551–600 | 5 | 0.07 |
| 601–650 | 5 | 0.07 |
- See Table and Table .
-
In the following histogram data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where both boundary values are included).
Figure \(\PageIndex{6}\): -
In the following histogram, the data values that fall on the right boundary are counted in the class interval, while values that fall on the left boundary are not counted (with the exception of the first interval where values on both boundaries are included).
Figure \(\PageIndex{7}\): -
Compare the two graphs:
-
Answers may vary. Possible answers include:
- Both graphs have a single peak.
- Both graphs use class intervals with width equal to $50.
-
Answers may vary. Possible answers include:
- The couples graph has a class interval with no values.
- It takes almost twice as many class intervals to display the data for couples.
- Answers may vary. Possible answers include: The graphs are more similar than different because the overall patterns for the graphs are the same.
-
Answers may vary. Possible answers include:
- Check student's solution.
-
Compare the graph for the Singles with the new graph for the Couples:
-
- Both graphs have a single peak.
- Both graphs display 6 class intervals.
- Both graphs show the same general pattern.
- Answers may vary. Possible answers include: Although the width of the class intervals for couples is double that of the class intervals for singles, the graphs are more similar than they are different.
-
- Answers may vary. Possible answers include: You are able to compare the graphs interval by interval. It is easier to compare the overall patterns with the new scale on the Couples graph. Because a couple represents two individuals, the new scale leads to a more accurate comparison.
- Answers may vary. Possible answers include: Based on the histograms, it seems that spending does not vary much from singles to individuals who are part of a couple. The overall patterns are the same. The range of spending for couples is approximately double the range for individuals.
Q 2.3.3
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows.
| # of movies | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
| 0 | 5 | ||
| 1 | 9 | ||
| 2 | 6 | ||
| 3 | 4 | ||
| 4 | 1 |
- Construct a histogram of the data.
- Complete the columns of the chart.
Use the following information to answer the next two exercises: Suppose one hundred eleven people who shopped in a special t-shirt store were asked the number of t-shirts they own costing more than $19 each.
Figure <IP_ADDRESS>.Q 2.3.4
The percentage of people who own at most three t-shirts costing more than $19 each is approximately:
- 21
- 59
- 41
- Cannot be determined
S 2.3.4
c
Q 2.3.5
If the data were collected by asking the first 111 people who entered the store, then the type of sampling is:
- cluster
- simple random
- stratified
- convenience
Q 2.3.6
Following are the 2010 obesity rates by U.S. states and Washington, DC.
| State | Percent (%) | State | Percent (%) | State | Percent (%) |
|---|---|---|---|---|---|
| Alabama | 32.2 | Kentucky | 31.3 | North Dakota | 27.2 |
| Alaska | 24.5 | Louisiana | 31.0 | Ohio | 29.2 |
| Arizona | 24.3 | Maine | 26.8 | Oklahoma | 30.4 |
| Arkansas | 30.1 | Maryland | 27.1 | Oregon | 26.8 |
| California | 24.0 | Massachusetts | 23.0 | Pennsylvania | 28.6 |
| Colorado | 21.0 | Michigan | 30.9 | Rhode Island | 25.5 |
| Connecticut | 22.5 | Minnesota | 24.8 | South Carolina | 31.5 |
| Delaware | 28.0 | Mississippi | 34.0 | South Dakota | 27.3 |
| Washington, DC | 22.2 | Missouri | 30.5 | Tennessee | 30.8 |
| Florida | 26.6 | Montana | 23.0 | Texas | 31.0 |
| Georgia | 29.6 | Nebraska | 26.9 | Utah | 22.5 |
| Hawaii | 22.7 | Nevada | 22.4 | Vermont | 23.2 |
| Idaho | 26.5 | New Hampshire | 25.0 | Virginia | 26.0 |
| Illinois | 28.2 | New Jersey | 23.8 | Washington | 25.5 |
| Indiana | 29.6 | New Mexico | 25.1 | West Virginia | 32.5 |
| Iowa | 28.4 | New York | 23.9 | Wisconsin | 26.3 |
| Kansas | 29.4 | North Carolina | 27.8 | Wyoming | 25.1 |
Construct a bar graph of obesity rates of your state and the four states closest to your state. Hint: Label the \(x\)-axis with the states.
S 2.3.7
Answers will vary.
Exercise 2.3.6
Sixty-five randomly selected car salespersons were asked the number of cars they generally sell in one week. Fourteen people answered that they generally sell three cars; nineteen generally sell four cars; twelve generally sell five cars; nine generally sell six cars; eleven generally sell seven cars. Complete the table.
| Data Value (# cars) | Frequency | Relative Frequency | Cumulative Relative Frequency |
|---|---|---|---|
Exercise 2.3.8
What does the relative frequency column in Table sum to? Why?
Exercise 2.3.9
What is the difference between relative frequency and frequency for each data value in Table ?
Answer
The relative frequency shows the proportion of data points that have each value. The frequency tells the number of data points that have each value.
Exercise 2.3.10
What is the difference between cumulative relative frequency and relative frequency for each data value?
Exercise 2.3.11
To construct the histogram for the data in Table , determine appropriate minimum and maximum x and y values and the scaling. Sketch the histogram. Label the horizontal and vertical axes with words. Include numerical scaling.
Figure \(\PageIndex{9}\):Answer
Answers will vary. One possible histogram is shown:
Figure \(\PageIndex{10}\):Exercise 2.3.12
Construct a frequency polygon for the following:
-
Pulse Rates for Women Frequency 60–69 12 70–79 14 80–89 11 90–99 1 100–109 1 110–119 0 120–129 1 -
Actual Speed in a 30 MPH Zone Frequency 42–45 25 46–49 14 50–53 7 54–57 3 58–61 1 -
Tar (mg) in Nonfiltered Cigarettes Frequency 10–13 1 14–17 0 18–21 15 22–25 7 26–29 2
Exercise 2.3.13
Construct a frequency polygon from the frequency distribution for the 50 highest ranked countries for depth of hunger.
| Depth of Hunger | Frequency |
|---|---|
| 230–259 | 21 |
| 260–289 | 13 |
| 290–319 | 5 |
| 320–349 | 7 |
| 350–379 | 1 |
| 380–409 | 1 |
| 410–439 | 1 |
Answer
Find the midpoint for each class. These will be graphed on the x -axis. The frequency values will be graphed on the y -axis values.
Figure \(\PageIndex{11}\):Exercise 2.3.14
Use the two frequency tables to compare the life expectancy of men and women from 20 randomly selected countries. Include an overlayed frequency polygon and discuss the shapes of the distributions, the center, the spread, and any outliers. What can we conclude about the life expectancy of women compared to men?
| Life Expectancy at Birth – Women | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 3 |
| 77–83 | 8 |
| 84–90 | 2 |
| Life Expectancy at Birth – Men | Frequency |
|---|---|
| 49–55 | 3 |
| 56–62 | 3 |
| 63–69 | 1 |
| 70–76 | 1 |
| 77–83 | 7 |
| 84–90 | 5 |
Exercise 2.3.15
Construct a times series graph for (a) the number of male births, (b) the number of female births, and (c) the total number of births.
| Sex/Year | 1855 | 1856 | 1857 | 1858 | 1859 | 1860 | 1861 |
| Female | 45,545 | 49,582 | 50,257 | 50,324 | 51,915 | 51,220 | 52,403 |
| Male | 47,804 | 52,239 | 53,158 | 53,694 | 54,628 | 54,409 | 54,606 |
| Total | 93,349 | 101,821 | 103,415 | 104,018 | 106,543 | 105,629 | 107,009 |
| Sex/Year | 1862 | 1863 | 1864 | 1865 | 1866 | 1867 | 1868 | 1869 |
| Female | 51,812 | 53,115 | 54,959 | 54,850 | 55,307 | 55,527 | 56,292 | 55,033 |
| Male | 55,257 | 56,226 | 57,374 | 58,220 | 58,360 | 58,517 | 59,222 | 58,321 |
| Total | 107,069 | 109,341 | 112,333 | 113,070 | 113,667 | 114,044 | 115,514 | 113,354 |
| Sex/Year | 1871 | 1870 | 1872 | 1871 | 1872 | 1827 | 1874 | 1875 |
| Female | 56,099 | 56,431 | 57,472 | 56,099 | 57,472 | 58,233 | 60,109 | 60,146 |
| Male | 60,029 | 58,959 | 61,293 | 60,029 | 61,293 | 61,467 | 63,602 | 63,432 |
| Total | 116,128 | 115,390 | 118,765 | 116,128 | 118,765 | 119,700 | 123,711 | 123,578 |
Answer
Figure \(\PageIndex{12}\):Exercise 2.3.16
The following data sets list full time police per 100,000 citizens along with homicides per 100,000 citizens for the city of Detroit, Michigan during the period from 1961 to 1973.
| Year | 1961 | 1962 | 1963 | 1964 | 1965 | 1966 | 1967 |
| Police | 260.35 | 269.8 | 272.04 | 272.96 | 272.51 | 261.34 | 268.89 |
| Homicides | 8.6 | 8.9 | 8.52 | 8.89 | 13.07 | 14.57 | 21.36 |
| Year | 1968 | 1969 | 1970 | 1971 | 1972 | 1973 |
| Police | 295.99 | 319.87 | 341.43 | 356.59 | 376.69 | 390.19 |
| Homicides | 28.03 | 31.49 | 37.39 | 46.26 | 47.24 | 52.33 |
- Construct a double time series graph using a common x -axis for both sets of data.
- Which variable increased the fastest? Explain.
- Did Detroit’s increase in police officers have an impact on the murder rate? Explain.
2.4: Measures of the Location of the Data
Q 2.4.1
The median age for U.S. blacks currently is 30.9 years; for U.S. whites it is 42.3 years.
- Based upon this information, give two reasons why the black median age could be lower than the white median age.
- Does the lower median age for blacks necessarily mean that blacks die younger than whites? Why or why not?
- How might it be possible for blacks and whites to die at approximately the same age, but for the median age for whites to be higher?
Q 2.4.2
Six hundred adult Americans were asked by telephone poll, "What do you think constitutes a middle-class income?" The results are in Table . Also, include left endpoint, but not the right endpoint.
| Salary ($) | Relative Frequency |
|---|---|
| < 20,000 | 0.02 |
| 20,000–25,000 | 0.09 |
| 25,000–30,000 | 0.19 |
| 30,000–40,000 | 0.26 |
| 40,000–50,000 | 0.18 |
| 50,000–75,000 | 0.17 |
| 75,000–99,999 | 0.02 |
| 100,000+ | 0.01 |
- What percentage of the survey answered "not sure"?
- What percentage think that middle-class is from $25,000 to $50,000?
-
Construct a histogram of the data.
- Should all bars have the same width, based on the data? Why or why not?
- How should the <20,000 and the 100,000+ intervals be handled? Why?
- Find the 40 th and 80 th percentiles
- Construct a bar graph of the data
S 2.4.2
- \(1 - (0.02 + 0.09 + 0.19 + 0.26 + 0.18 + 0.17 + 0.02 + 0.01) = 0.06\)
- \(0.19 + 0.26 + 0.18 = 0.63\)
- Check student’s solution.
-
40
th
percentile will fall between 30,000 and 40,000
80 th percentile will fall between 50,000 and 75,000
- Check student’s solution.
Q 2.4.3
Given the following box plot:
Figure \(\PageIndex{13}\):- which quarter has the smallest spread of data? What is that spread?
- which quarter has the largest spread of data? What is that spread?
- find the interquartile range ( IQR ).
- are there more data in the interval 5–10 or in the interval 10–13? How do you know this?
-
which interval has the fewest data in it? How do you know this?
- 0–2
- 2–4
- 10–12
- 12–13
- need more information
Q 2.4.4
The following box plot shows the U.S. population for 1990, the latest available year.
Figure \(\PageIndex{14}\):- Are there fewer or more children (age 17 and under) than senior citizens (age 65 and over)? How do you know?
- 12.6% are age 65 and over. Approximately what percentage of the population are working age adults (above age 17 to age 65)?
S 2.4.4
- more children; the left whisker shows that 25% of the population are children 17 and younger. The right whisker shows that 25% of the population are adults 50 and older, so adults 65 and over represent less than 25%.
- 62.4%
2.5: Box Plots
Q 2.5.1
In a survey of 20-year-olds in China, Germany, and the United States, people were asked the number of foreign countries they had visited in their lifetime. The following box plots display the results.
Figure \(\PageIndex{15}\):- In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected.
- Have more Americans or more Germans surveyed been to over eight foreign countries?
- Compare the three box plots. What do they imply about the foreign travel of 20-year-old residents of the three countries when compared to each other?
Q 2.5.2
Given the following box plot, answer the questions.
Figure \(\PageIndex{16}\):- Think of an example (in words) where the data might fit into the above box plot. In 2–5 sentences, write down the example.
- What does it mean to have the first and second quartiles so close together, while the second to third quartiles are far apart?
S 2.5.2
- Answers will vary. Possible answer: State University conducted a survey to see how involved its students are in community service. The box plot shows the number of community service hours logged by participants over the past year.
- Because the first and second quartiles are close, the data in this quarter is very similar. There is not much variation in the values. The data in the third quarter is much more variable, or spread out. This is clear because the second quartile is so far away from the third quartile.
Q 2.5.3
Given the following box plots, answer the questions.
Figure \(\PageIndex{17}\):-
In complete sentences, explain why each statement is false.
- Data 1 has more data values above two than Data 2 has above two.
- The data sets cannot have the same mode.
- For Data 1 , there are more data values below four than there are above four.
- For which group, Data 1 or Data 2, is the value of “7” more likely to be an outlier? Explain why in complete sentences.
Q 2.5.4
A survey was conducted of 130 purchasers of new BMW 3 series cars, 130 purchasers of new BMW 5 series cars, and 130 purchasers of new BMW 7 series cars. In it, people were asked the age they were when they purchased their car. The following box plots display the results.
Figure \(\PageIndex{18}\):- In complete sentences, describe what the shape of each box plot implies about the distribution of the data collected for that car series.
- Which group is most likely to have an outlier? Explain how you determined that.
- Compare the three box plots. What do they imply about the age of purchasing a BMW from the series when compared to each other?
- Look at the BMW 5 series. Which quarter has the smallest spread of data? What is the spread?
- Look at the BMW 5 series. Which quarter has the largest spread of data? What is the spread?
- Look at the BMW 5 series. Estimate the interquartile range (IQR).
- Look at the BMW 5 series. Are there more data in the interval 31 to 38 or in the interval 45 to 55? How do you know this?
-
Look at the BMW 5 series. Which interval has the fewest data in it? How do you know this?
- 31–35
- 38–41
- 41–64
S 2.5.4
- Each box plot is spread out more in the greater values. Each plot is skewed to the right, so the ages of the top 50% of buyers are more variable than the ages of the lower 50%.
- The BMW 3 series is most likely to have an outlier. It has the longest whisker.
- Comparing the median ages, younger people tend to buy the BMW 3 series, while older people tend to buy the BMW 7 series. However, this is not a rule, because there is so much variability in each data set.
- The second quarter has the smallest spread. There seems to be only a three-year difference between the first quartile and the median.
- The third quarter has the largest spread. There seems to be approximately a 14-year difference between the median and the third quartile.
- IQR ~ 17 years
- There is not enough information to tell. Each interval lies within a quarter, so we cannot tell exactly where the data in that quarter is concentrated.
- The interval from 31 to 35 years has the fewest data values. Twenty-five percent of the values fall in the interval 38 to 41, and 25% fall between 41 and 64. Since 25% of values fall between 31 and 38, we know that fewer than 25% fall between 31 and 35.
Q 2.5.5
Twenty-five randomly selected students were asked the number of movies they watched the previous week. The results are as follows:
| # of movies | Frequency |
|---|---|
| 0 | 5 |
| 1 | 9 |
| 2 | 6 |
| 3 | 4 |
| 4 | 1 |
Construct a box plot of the data.
2.6: Measures of the Center of the Data
Q 2.6.1
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in the following table.
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
- What is the best estimate of the average obesity percentage for these countries?
- The United States has an average obesity rate of 33.9%. Is this rate above average or below?
- How does the United States compare to other countries?
Q 2.6.2
Table gives the percent of children under five considered to be underweight. What is the best estimate for the mean percentage of underweight children?
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
S 2.6.2
The mean percentage, \(\bar{x} = \frac{1328.65}{50} = 26.75\)
2.7: Skewness and the Mean, Median, and Mode
Q 2.7.1
The median age of the U.S. population in 1980 was 30.0 years. In 1991, the median age was 33.1 years.
- What does it mean for the median age to rise?
- Give two reasons why the median age could rise.
- For the median age to rise, is the actual number of children less in 1991 than it was in 1980? Why or why not?
2.8: Measures of the Spread of the Data
Use the following information to answer the next nine exercises: The population parameters below describe the full-time equivalent number of students (FTES) each year at Lake Tahoe Community College from 1976–1977 through 2004–2005.
- \(\mu = 1000\) FTES
- median = 1,014 FTES
- \(\sigma = 474\) FTES
- first quartile = 528.5 FTES
- third quartile = 1,447.5 FTES
- \(n = 29\) years
Q 2.8.1
A sample of 11 years is taken. About how many are expected to have a FTES of 1014 or above? Explain how you determined your answer.
S 2.8.1
The median value is the middle value in the ordered list of data values. The median value of a set of 11 will be the 6th number in order. Six years will have totals at or below the median.
Q 2.8.2
75% of all years have an FTES:
- at or below: _____
- at or above: _____
Q 2.8.3
The population standard deviation = _____
S 2.8.3
474 FTES
Q 2.8.4
What percent of the FTES were from 528.5 to 1447.5? How do you know?
Q 2.8.5
What is the IQR ? What does the IQR represent?
S 2.8.5
919
Q 2.8.6
How many standard deviations away from the mean is the median?
Additional Information: The population FTES for 2005–2006 through 2010–2011 was given in an updated report. The data are reported here.
| Year | 2005–06 | 2006–07 | 2007–08 | 2008–09 | 2009–10 | 2010–11 |
| Total FTES | 1,585 | 1,690 | 1,735 | 1,935 | 2,021 | 1,890 |
Q 2.8.7
Calculate the mean, median, standard deviation, the first quartile, the third quartile and the IQR . Round to one decimal place.
S 2.8.7
- mean = 1,809.3
- median = 1,812.5
- standard deviation = 151.2
- first quartile = 1,690
- third quartile = 1,935
- IQR = 245
Q 2.8.8
Construct a box plot for the FTES for 2005–2006 through 2010–2011 and a box plot for the FTES for 1976–1977 through 2004–2005.
Q 2.8.9
Compare the IQR for the FTES for 1976–77 through 2004–2005 with the IQR for the FTES for 2005-2006 through 2010–2011. Why do you suppose the IQR s are so different?
S 2.8.10
Hint: Think about the number of years covered by each time period and what happened to higher education during those periods.
Q 2.8.11
Three students were applying to the same graduate school. They came from schools with different grading systems. Which student had the best GPA when compared to other students at his school? Explain how you determined your answer.
| Student | GPA | School Average GPA | School Standard Deviation |
|---|---|---|---|
| Thuy | 2.7 | 3.2 | 0.8 |
| Vichet | 87 | 75 | 20 |
| Kamala | 8.6 | 8 | 0.4 |
Q 2.8.12
A music school has budgeted to purchase three musical instruments. They plan to purchase a piano costing $3,000, a guitar costing $550, and a drum set costing $600. The mean cost for a piano is $4,000 with a standard deviation of $2,500. The mean cost for a guitar is $500 with a standard deviation of $200. The mean cost for drums is $700 with a standard deviation of $100. Which cost is the lowest, when compared to other instruments of the same type? Which cost is the highest when compared to other instruments of the same type. Justify your answer.
S 2.8.12
For pianos, the cost of the piano is 0.4 standard deviations BELOW the mean. For guitars, the cost of the guitar is 0.25 standard deviations ABOVE the mean. For drums, the cost of the drum set is 1.0 standard deviations BELOW the mean. Of the three, the drums cost the lowest in comparison to the cost of other instruments of the same type. The guitar costs the most in comparison to the cost of other instruments of the same type.
Q 2.8.13
An elementary school class ran one mile with a mean of 11 minutes and a standard deviation of three minutes. Rachel, a student in the class, ran one mile in eight minutes. A junior high school class ran one mile with a mean of nine minutes and a standard deviation of two minutes. Kenji, a student in the class, ran 1 mile in 8.5 minutes. A high school class ran one mile with a mean of seven minutes and a standard deviation of four minutes. Nedda, a student in the class, ran one mile in eight minutes.
- Why is Kenji considered a better runner than Nedda, even though Nedda ran faster than he?
- Who is the fastest runner with respect to his or her class? Explain why.
Q 2.8.14
The most obese countries in the world have obesity rates that range from 11.4% to 74.6%. This data is summarized in Table 14 .
| Percent of Population Obese | Number of Countries |
|---|---|
| 11.4–20.45 | 29 |
| 20.45–29.45 | 13 |
| 29.45–38.45 | 4 |
| 38.45–47.45 | 0 |
| 47.45–56.45 | 2 |
| 56.45–65.45 | 1 |
| 65.45–74.45 | 0 |
| 74.45–83.45 | 1 |
What is the best estimate of the average obesity percentage for these countries? What is the standard deviation for the listed obesity rates? The United States has an average obesity rate of 33.9%. Is this rate above average or below? How “unusual” is the United States’ obesity rate compared to the average rate? Explain.
S 2.8.14
- \(\bar{x} = 23.32\)
- Using the LibreTexts calculator, we obtain a standard deviation of: \(s_{x} = 12.95\).
- The obesity rate of the United States is 10.58% higher than the average obesity rate.
- Since the standard deviation is 12.95, we see that \(23.32 + 12.95 = 36.27\) is the obesity percentage that is one standard deviation from the mean. The United States obesity rate is slightly less than one standard deviation from the mean. Therefore, we can assume that the United States, while 34% obese, does not have an unusually high percentage of obese people.
Q 2.8.15
Table gives the percent of children under five considered to be underweight.
| Percent of Underweight Children | Number of Countries |
|---|---|
| 16–21.45 | 23 |
| 21.45–26.9 | 4 |
| 26.9–32.35 | 9 |
| 32.35–37.8 | 7 |
| 37.8–43.25 | 6 |
| 43.25–48.7 | 1 |
What is the best estimate for the mean percentage of underweight children? What is the standard deviation? Which interval(s) could be considered unusual? Explain.


8.2.3: Nitrogen Effects on Bacterial Growth
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Learning Objectives
- Distinguish between component I and II of the nitrogenase enzyme and its role in biological nitrogen fixation
Biological nitrogen fixation (BNF) occurs when atmospheric nitrogen is converted to ammonia by an enzyme called nitrogenase. The reaction for BNF is:
\[\ce{N2 + 8 H^{+} + 8 e^{−} → 2 NH3 + H2}.\]
This type of reaction results in N 2 gaining electrons (see above equation) and is thus termed a reduction reaction. The exact mechanism of catalysis is unknown due to the technical difficulties biochemists have in actually visualizing this reaction in vitro, so the exact sequence of the steps of this reaction are not completely understood. Despite this, a great deal is known of the process. While the equilibrium formation of ammonia from molecular hydrogen and nitrogen has an overall negative enthalpy of reaction (i.e. it gives off energy), the energy barrier to activation is very high without the assistance of catalysis, which is done by nitrogenases. The enzymatic reduction of N 2 to ammonia therefore requires an input of chemical energy, released from ATP hydrolysis, to overcome the activation energy barrier.
Nitrogenase is made up of two soluble proteins: component I and II. Component I known as MoFe protein or nitrogenase contains 2 Mo atoms, 28 to 34 Fe atoms, and 26 to 28 acid-labile sulfides, also known as a iron-molybdenum cofactor (FeMoco). Component I is composed of two copies each of two subunits (α and β); each subunit’s stability depends on the other in vivo. Component II known as Fe protein or nitrogenase reductase is composed of two copies of a single subunit. This protein has four non-heme Fe atoms and four acid-labile sulfides (4Fe-4S). Substrate binding and reduction takes place on component I, which binds to ATP and ferredoxin or flavodoxin proteins (Fdx or Fld) (see step B).
The hydrolysis of ATP supplies the energy for the reaction while the Fdx/Fld proteins supply the electrons. Note this is a reduction reaction which means that electrons must be added to the N 2 to reduce it to NH 4 . Thus, the role of component II is to supply electrons, one at a time to component I. ATP is not hydrolyzed to ADP until component II transfers an electron to component I (see step C and D). 21-25 ATPs are required for each N 2 fixed. The association of nitrogenase component I and II and later dissociation occurs several times to allow the fixation of one N 2 molecule (see step B and D).
Nitrogenase ultimately bonds each atom of nitrogen to three hydrogen atoms to form ammonia (NH 3 ). The nitrogenase reaction additionally produces molecular hydrogen as a side product, which is of special interest for people trying to produce H 2 as an alternative energy source to fossil fuels.
Key Points
- Nitrogen fixation does result in the release of energy, but the activation of this reaction takes energy in the form of ATP hydrolysis.
- Nitrogenases are metalloenzymes, which are proteins that have metalic molecules as subunits.
- While a great deal is known about how nitrogenases reduce nitrogen, some steps are unknown.
Key Terms
- sulfide : Any compound of sulfur and a metal or other electropositive element or group.
- reduction : A reaction in which electrons are gained and valence is reduced; often by the removal of oxygen or the addition of hydrogen.
- enthalpy : In thermodynamics, a measure of the heat content of a chemical or physical system.


5.3: Decimal Operations (Part 1)
- Add and subtract decimals
- Multiply decimals
- Divide decimals
- Use decimals in money applications
Before you get started, take this readiness quiz.
- Simplify \(\dfrac{70}{100}\). If you missed this problem, review Example 4.3.1 .
- Multiply \(\dfrac{3}{10} \cdot \dfrac{9}{10}\). If you missed this problem, review Example 4.3.7 .
- Divide −36 ÷ (−9). If you missed this problem, review Example 3.7.3 .
Add and Subtract Decimals
Let’s take one more look at the lunch order from the start of Decimals , this time noticing how the numbers were added together.
\[\begin{split} & $3.45 \quad Sandwich \\ & $1.25 \quad Water \\ + & $0.33 \quad Tax \\ \hline & $5.03 \quad Total \end{split}\]
All three items (sandwich, water, tax) were priced in dollars and cents, so we lined up the dollars under the dollars and the cents under the cents, with the decimal points lined up between them. Then we just added each column, as if we were adding whole numbers. By lining up decimals this way, we can add or subtract the corresponding place values just as we did with whole numbers.
Step 1. Write the numbers vertically so the decimal points line up.
Step 2. Use zeros as place holders, as needed.
Step 3. Add or subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers.
Add: 3.7 + 12.4.
Solution
| Write the numbers vertically so the decimal points line up. | $$\begin{split} 3.&7 \\ + 12.&4 \\ \hline \end{split}$$ |
| Place holders are not needed since both numbers have the same number of decimal places. | |
| Add the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. | $$\begin{split} \stackrel{1}{3}.&7 \\ + 12.&4 \\ \hline 16.&1 \end{split}$$ |
Add: 5.7 + 11.9.
- Answer
-
\(17.6\)
Add: 18.32 + 14.79.
- Answer
-
\(13.11\)
Add: 23.5 + 41.38.
Solution
| Write the numbers vertically so the decimal points line up. | $$\begin{split} 23.&5 \\ + 41.&38 \\ \hline \end{split}$$ |
| Place 0 as a place holder after the 5 in 23.5, so that both numbers have two decimal places. | $$\begin{split} 23.&5 \textcolor{red}{0} \\ + 41.&38 \\ \hline \end{split}$$ |
| Add the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. | $$\begin{split} 23.&50 \\ + 41.&38 \\ \hline 64.&88 \end{split}$$ |
Add: 4.8 + 11.69.
- Answer
-
\(16.49\)
Add: 5.123 + 18.47.
- Answer
-
\(23.593\)
How much change would you get if you handed the cashier a $20 bill for a $14.65 purchase? We will show the steps to calculate this in the next example.
Subtract: 20 − 14.65.
Solution
| Write the numbers vertically so the decimal points line up. Remember 20 is a whole number, so place the decimal point after the 0. | $$\begin{split} 20.& \\ - 14.&65 \\ \hline \end{split}$$ |
| Place two zeros after the decimal point in 20, as place holders so that both numbers have two decimal places. |
\[\begin{split} 20.& \textcolor{red}{00} \\ - 14.&65 \\ \hline \end{split}\] |
| Subtract the numbers as if they were whole numbers. Then place the decimal in the answer under the decimal points in the given numbers. | $$\begin{split} \stackrel{1}{\cancel{2}} \stackrel{\stackrel{9}{\cancel{10}}}{\cancel{0}} &.\stackrel{\stackrel{9}{\cancel{10}}}{\cancel{0}} \stackrel{\stackrel{9}{\cancel{10}}}{\cancel{0}} \\ - 1\; \; 4\; \; &.\; 6\; \; 5 \\ \hline 5\; \; &.\; 3\; \; 5\end{split}$$ |
Subtract: 10 − 9.58.
- Answer
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\(0.42\)
Subtract: 50 − 37.42.
- Answer
-
\(12.58\)
Subtract: 2.51 − 7.4.
Solution
If we subtract 7.4 from 2.51, the answer will be negative since 7.4 > 2.51. To subtract easily, we can subtract 2.51 from 7.4. Then we will place the negative sign in the result.
| Write the numbers vertically so the decimal points line up. | $$\begin{split} 7.&4 \\ - 2.&51 \\ \hline \end{split}$$ |
| Place zero after the 4 in 7.4 as a place holder, so that both numbers have two decimal places. | $$\begin{split} 7.&4 \textcolor{red}{0} \\ - 2.&51 \\ \hline \end{split}$$ |
| Subtract and place the decimal in the answer. | $$\begin{split} 7.&40 \\ - 2.&51 \\ \hline 4.&89 \end{split}$$ |
| Remember that we are really subtracting 2.51 − 7.4 so the answer is negative. | 2.51 − 7.4 = − 4.89 |
Subtract: 4.77 − 6.3.
- Answer
-
\(-1.53\)
Subtract: 8.12 − 11.7.
- Answer
-
\(-3.58\)
Multiply Decimals
Multiplying decimals is very much like multiplying whole numbers—we just have to determine where to place the decimal point. The procedure for multiplying decimals will make sense if we first review multiplying fractions.
Do you remember how to multiply fractions? To multiply fractions, you multiply the numerators and then multiply the denominators. So let’s see what we would get as the product of decimals by converting them to fractions first. We will do two examples side-by-side in Table 5.22. Look for a pattern.
| A | B | |
|---|---|---|
| (0.3)(0.7) | (0.2)(0.46) | |
| Convert to fractions. | $$\left(\dfrac{3}{10}\right) \left(\dfrac{7}{10}\right)$$ | $$\left(\dfrac{2}{10}\right) \left(\dfrac{46}{100}\right)$$ |
| Multiply. | $$\dfrac{21}{100}$$ | $$\dfrac{92}{1000}$$ |
| Convert back to decimals | 0.21 | 0.092 |
There is a pattern that we can use. In A, we multiplied two numbers that each had one decimal place, and the product had two decimal places. In B, we multiplied a number with one decimal place by a number with two decimal places, and the product had three decimal places.
How many decimal places would you expect for the product of (0.01)(0.004)? If you said “five”, you recognized the pattern. When we multiply two numbers with decimals, we count all the decimal places in the factors—in this case two plus three—to get the number of decimal places in the product—in this case five.
Once we know how to determine the number of digits after the decimal point, we can multiply decimal numbers without converting them to fractions first. The number of decimal places in the product is the sum of the number of decimal places in the factors.
The rules for multiplying positive and negative numbers apply to decimals, too, of course.
When multiplying two numbers,
- if their signs are the same, the product is positive.
- if their signs are different, the product is negative.
When you multiply signed decimals, first determine the sign of the product and then multiply as if the numbers were both positive. Finally, write the product with the appropriate sign.
Step 1. Determine the sign of the product.
Step 2. Write the numbers in vertical format, lining up the numbers on the right.
Step 3. Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points.
Step 4. Place the decimal point. The number of decimal places in the product is the sum of the number of decimal places in the factors. If needed, use zeros as placeholders.
Step 5. Write the product with the appropriate sign.
Multiply: (3.9)(4.075).
Solution
| Determine the sign of the product. The signs are the same. | The product will be positive. |
| Write the numbers in vertical format, lining up the numbers on the right. | $$\begin{split} 4.07&5 \\ \times 3.&9 \\ \hline \end{split}$$ |
| Multiply the numbers as if they were whole numbers, temporarily ignoring the decimal points. | $$\begin{split} 4.07&5 \\ \times 3.&9 \\ \hline 3667&5 \\ 12225&\; \\ \hline 15892&5 \end{split}$$ |
| Place the decimal point. Add the number of decimal places in the factors (1 + 3). Place the decimal point 4 places from the right. | $$\begin{split} 4.07&5 \quad \textcolor{blue}{3\; places} \\ \times 3.&9 \quad \textcolor{blue}{1\; place}\\ \hline 3667&5 \\ 12225&\; \\ \hline 15892&5 \quad \textcolor{blue}{4\; places} \end{split}$$ |
| The product is positive. | (3.9)(4.075) = 15.8925 |
Multiply: 4.5(6.107).
- Answer
-
\(27.4815\)
Multiply: 10.79(8.12).
- Answer
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\(87.6148\)
Multiply: (−8.2)(5.19).
Solution
| The signs are different. | The product will be negative. |
| Write in vertical format, lining up the numbers on the right. | $$\begin{split} 5.&19 \\ \times 8.&2 \\ \hline \end{split}$$ |
| Multiply. | $$\begin{split} 5.&19 \\ \times 8.&2 \\ \hline 10&38 \\ 415&2\; \\ \hline 425&58 \end{split}$$ |
| $$\begin{split} 5.&19 \\ \times 8.&2 \\ \hline 10&38 \\ 415&2\; \\ \hline 42.5&58 \end{split}$$ | |
| The product is negative. | (−8.2)(5.19) = −42.558 |
Multiply: (4.63)(−2.9).
- Answer
-
\(-13.427\)
Multiply: (−7.78)(4.9).
- Answer
-
\(-38.122\)
In the next example, we’ll need to add several placeholder zeros to properly place the decimal point.
Multiply: (0.03)(0.045).
Solution
| The product is positive. | (0.03)(0.045) |
| Write in vertical format, lining up the numbers on the right. | $$\begin{split} 0.04&5 \\ \times 0.0&3 \\ \hline \end{split}$$ |
| Multiply. | $$\begin{split} 0.04&5 \\ \times 0.0&3 \\ \hline 13&5 \end{split}$$ |
|
Add zeros as needed to get the 5 places. |
|
| The product is positive. | (0.03)(0.045) = 0.00135 |
Multiply: (0.04)(0.087).
- Answer
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\(0.00348\)
Multiply: (0.09)(0.067).
- Answer
-
\(0.00603\)
Multiply by Powers of 10
In many fields, especially in the sciences, it is common to multiply decimals by powers of 10. Let’s see what happens when we multiply 1.9436 by some powers of 10.
Look at the results without the final zeros. Do you notice a pattern?
\[\begin{split} 1.9436(10) & = 19.436 \\ 1.9436(100) & = 194.36 \\ 1.9436(1000) & = 1943.6 \end{split}\]
The number of places that the decimal point moved is the same as the number of zeros in the power of ten. Table 5.26 summarizes the results.
| Multiply by | Number of zeros | Number of places decimal point moves |
|---|---|---|
| 10 | 1 | 1 place to the right |
| 100 | 2 | 2 places to the right |
| 1,000 | 3 | 3 places to the right |
| 10,000 | 4 | 4 places to the right |
We can use this pattern as a shortcut to multiply by powers of ten instead of multiplying using the vertical format. We can count the zeros in the power of 10 and then move the decimal point that same of places to the right. So, for example, to multiply 45.86 by 100, move the decimal point 2 places to the right.
Sometimes when we need to move the decimal point, there are not enough decimal places. In that case, we use zeros as placeholders. For example, let’s multiply 2.4 by 100. We need to move the decimal point 2 places to the right. Since there is only one digit to the right of the decimal point, we must write a 0 in the hundredths place.
Step 1. Move the decimal point to the right the same number of places as the number of zeros in the power of 10.
Step 2. Write zeros at the end of the number as placeholders if needed.
Multiply 5.63 by factors of (a) 10 (b) 100 (c) 1000.
Solution
By looking at the number of zeros in the multiple of ten, we see the number of places we need to move the decimal to the right.
(a) 5.63(10)
| There is 1 zero in 10, so move the decimal point 1 place to the right. | |
| 56.3 |
(b) 5.63(100)
| There are 2 zeros in 100, so move the decimal point 2 places to the right. | |
| 563 |
(c) 5.63(1000)
| There are 3 zeros in 1000, so move the decimal point 3 places to the right. | |
| A zero must be added at the end. | 5,630 |
Multiply 2.58 by factors of (a) 10 (b) 100 (c) 1000.
- Answer a
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\(25.8\)
- Answer b
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\(258\)
- Answer c
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\(2,580\)
Multiply 14.2 by factors of (a) 10 (b) 100 (c) 1000.
- Answer a
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\(142\)
- Answer b
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\(1,420\)
- Answer c
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\(14,200\)
Contributors and Attributions
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Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at<EMAIL_ADDRESS>."


1.11: Pre- Mid- and Post-Test-Taking Strategies
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Phyllis Nissila
Pre-Test Strategies
Q: When should you start preparing for the first test? Circle…
- The night before.
- The week prior.
- The first day of classes.
If you answered “3. The first day of classes,” you are correct. If you circled all three, you are also correct. Preparing to pass tests is something that begins when learning begins and continues all the way through to the final exam.
Many students, however, don’t start thinking about test taking, whether weekly exams, mid-terms, or finals, until the day before when they engage in an all-nighter, or cramming. From the previous unit on memory, you might recall that the brain can only process an average of 5-7 new pieces of information at a time. Additionally, unless memory devices are used to aid memory and to cement information into long term memory (or at least until the test is over tomorrow!) chances are slim that students who cram will effectively learn and remember the information.
Additionally, a lot of students are unaware of the many strategies available to help with the test-taking experience before, during, and after. For starters, take a look at what has helped you so far.
EXERCISE 30-1
Pre-Test Taking Strategies
PART A:
Put a check mark next to the pre-test strategies you already employ.
____ Organize your notebook and other class materials the first week of classes.
____ Maintain your organized materials throughout the term.
____ Take notes on key points from lectures and other materials.
____ Make sure you understand the information as you go along.
____ Access your instructor’s help and the help of a study group, as needed.
____ Organize a study group, if desired.
____ Create study tools such as flashcards, graphic organizers, etc. as study aids.
____ Complete all homework assignments on time.
____ Review likely test items several times beforehand.
____ Ask your instructor what items are likely to be covered on the test.
____ Ask your instructor if she or he can provide a study guide or practice test.
____ Ask your instructor if he/she gives partial credit for test items such as essays.
____ Maintain an active learner attitude.
____ Schedule extra study time in the days just prior to the test.
____ Gather all notes, handouts, and other materials needed before studying.
____ Review all notes, handouts, and other materials.
____ Organize your study area for maximum concentration and efficiency.
____ Create and use mnemonic devices to aid memory.
____ Put key terms, formulas, etc., on a single study sheet that can be quickly reviewed.
____ Schedule study times short enough (1-2 hours) so you do not get burned out.
____ Get plenty of sleep the night before.
____ Set a back-up alarm in case the first alarm doesn’t sound or you sleep through it.
____ Have a good breakfast with complex carbs and protein to see you through.
____ Show up 5-10 minutes early to get completely settled before the test begins.
____ Use the restroom beforehand to minimize distractions.
PART B
By reviewing the pre-test strategies, above, you have likely discovered new ideas to add to what you already use. Make a list of them.
Mid-Test Strategies
Here is a list of the most common–and useful–strategies to survive this ubiquitous college experience.
- Scan the test, first, to get the big picture of how many test items there are, what types there are (multiple choice, matching, essay, etc.), and the point values of each item or group of items.
- Determine which way you want to approach the test: Some students start with the easy questions first, that is, the ones they immediately know the answers to, saving the difficult ones for later, knowing they can spend the remaining time on them. Some students begin with the biggest-point items first, to make sure they get the most points.
- Determine a schedule that takes into consideration how long you have to test, and the types of questions on the test. Essay questions, for example, will require more time than multiple choice, or matching questions.
- Keep your eye on the clock.
- If you can mark on the test, put a check mark next to items you are not sure of just yet. It is easy to go back and find them to answer later on. You might just find help in other test questions covering similar information.
- Sit where you are most comfortable. That said, sitting near the front has a couple of advantages: You may be less distracted by other students. If a classmate comes up with a question for the instructor and there is an important clarification given, you will be better able to hear it and apply it, if needed.
- Wear ear plugs, if noise distracts you.
- You do NOT have to start with #1! It you are unsure of it, mark it to come back to later on.
- Bring water…this helps calm the nerves, for one, and water is also needed for optimum brain function.
- If permitted, get up and stretch (or stretch in your chair) from time to time to relieve tension and assist the blood to the brain!
- Remember to employ strategies to reduce test-taking anxiety (covered in the next lesson)
- If despite all of your best efforts to prepare for a test you just cannot remember the answer to a given item for multiple choice, matching, and/or true/false questions, employ one or more of the following educated guessing (also known as “educated selection”) techniques. By using these techniques, you have a better chance of selecting the correct answer. It is usually best to avoid selecting an extreme or all-inclusive answer (also known as 100% modifiers) such as “always,” and “never”. Choose, instead, words such as “usually,” “sometimes,” etc. (also known as in-between modifiers). If the answers are numbers, choose one of the middle numbers. If you have options such as “all of the above,” or “both A and B,” make sure each item is true before selecting those options. Choose the longest, or most inclusive, answer. Make sure to match the grammar of question and answer. For example, if the question indicates a plural answer, look for the plural answer. Regarding matching tests: count both sides to be matched. If there are more questions than answers, ask if you can use an answer more than once. Pay close attention to items that ask you to choose the “best” answer. This means one answer is better or more inclusive than a similar answer. Read all of the response options.
Post-Test Strategies
In addition to sighing that big sigh of relief, here are a few suggestions to help with future tests.
- If you don’t understand why you did not get an item right, ask the instructor. This is especially useful for quizzes that contain information that may be incorporated into more inclusive exams such as mid-terms and finals.
- Analyze your results to help you in the future. For example, see if most of your incorrect answers were small things such as failing to include the last step in a math item, or neglecting to double-check for simple errors in a short-answer or essay item. See where in the test you made the most errors: beginning, middle, or end. Also analyze which type of questions, true/false, multiple choice, essay, etc. And which topics were missed. This will help you pay closer attention to those sections in the future.
EXERCISE 30-2
Write a letter of advice to Chen incorporating 10 test-taking tips and strategies you think will help him.
Chen believes he is good at organization, and he usually is–for about the first two weeks of classes. He then becomes overwhelmed with all of the handouts and materials and tends to start slipping in the organization department. When it comes to tests, he worries that his notes might not cover all of the right topics and that he will not be able to remember all of the key terms and points–especially for his math class. During tests, he sometimes gets stuck on an item and tends to spend too much time there. He also sometimes changes answers but finds out later that his original selection was correct. Chen is also easily distracted by other students and noises which makes it hard for him to concentrate sometimes, and, unfortunately, he does admit to occasionally “cramming” the night before.
Licenses and Attributions:
CC licensed content, previously shared:
How to Learn Like a Pro! Authored by Phyllis Nissila. Located at:
https://openoregon.pressbooks.pub/co...st-strategies/
CC-BY Attribution.
Adaptions: Changed formatting, removed one exercise.
“Anxiety” image by HAMZA BUTT is licensed under CC BY 2.0


12: Newton's Second Law for Rigid Bodies Last updated Save as PDF Page ID 50610 Jacob Moore & Contributors Pennsylvania State University Mont Alto via Mechanics Map 12.1: Rigid Body Translation Introduction to analysis of rigid body motion; the difference between dynamics analysis and the static analysis covered in previous chapters. Analysis of translational motion for rigid bodies. Includes a worked example. 12.2: Fixed-Axis Rotation Analysis of fixed-axis rotation, both balanced and unbalanced, for rigid bodies. Includes worked examples. 12.3: Rigid-Body General Planar Motion Analysis of a rigid body undergoing general planar motion, incorporating both translation and rotation. Includes worked examples. 12.4: Multi-Body General Planar Motion Analysis of general planar motion for systems of multiple rigid bodies. Includes a worked example. 12.5: Chapter 12 Homework Problems


2.10: Scientific Management Theories
- Discuss classical scientific management theories
The foundation for modern human resource management was established in the early 1900s, with the emergence of scientific management principles. Influential classical scientific management theorists include Dr. Frederick Taylor as well as Frank Gilbreth and his wife, Dr. Lillian Gilbreth. The contributions of two additional thought leaders of the classical period, administrative theorist Henri Fayol and humanist Mary Parker Follett, will be discussed in subsequent sections.
Considered the “father of scientific management,” Dr. Taylor published his principles, also referred to as “Taylorism,” in a paper titled “The Principles of Scientific Management” in 1911. [1] In his introduction, Taylor observes that:
The search for better, for more competent men…was never more vigorous than it is now. And more than ever before is the demand for competent men in excess of the supply. What we are all looking for, however, is the ready-made, competent man; the man whom some one else has trained. It is only when we fully realize that our duty, as well as our opportunity, lies in systematically cooperating to train and to make this competent man, instead of in hunting for a man whom some one else has trained, that we shall be on the road to national efficiency.
Taylor’s argument was that the remedy to inefficiency was systematic management, rather than the search for an extraordinary human resource. Specifically, he sought to prove that “the best management is a true science,” and “to show that the fundamental principles of scientific management are applicable to human activities.” And, ultimately, that the correct application of these principles can yield exceptional improvements in efficiency.
In contrast to the labor abuses and hostility common in America’s early industrial age, Taylor believed that the purpose of management was to maximize both employer and employee prosperity. In terms of the individual employee, maximum prosperity means not only maximizing one’s wage but developing each man to his state of maximum efficiency. A mechanical engineer by training, Taylor designed a number of time and motion studies to determine the one best way to complete a specific task. Taylor distilled his research into four principles of scientific management:
- Apply the scientific method to analyze work and determine the most efficient way to perform every task.
- Match workers to jobs based on their capability and motivation and train them to work at maximum efficiency.
- Provide instruction and supervision to ensure that the “best” method is being used to complete each task.
- Divide work between managers and workers, with managers focusing on planning and training and workers focusing on production.
Dr. Lillian Gilbreth and her husband Frank were also pioneers in the field of scientific management. Like Taylor, the Gilbreth’s applied their engineering skills to identifying “the one best way” to perform a task. In addition to time and motion studies, the couple used film to identify opportunities to improve work processes. In his lifetime, Frank Gilbreth, the “father of motion study,” made over 250,000 feet of 35mm of motion picture films; for perspective, Motion Pictures in the Human Sciences’ website has a page with links to his films: The Original Films of Frank Gilbreth . Frank’s major contribution was to categorize human work as a number of individual motions, termed “Therbligs,” and then optimize these motions to improve efficiency—a process he applied to operations ranging from bricklaying to surgery. [2]
The couple, who had 12 children, also “mainstreamed home economics and domestic management.” [3] A psychologist as well as an industrial engineer, Lillian Gilbreth is credited with a number of inventions including shelves on the inside of refrigerator doors, the foot pedal on the trash can, wall light switches, and the linear kitchen layout. In a biography of Lillian Gilbreth, The American Society of Mechanical Engineer notes that she is referred to “the mother of modern management” and “was awarded the prestigious Hoover Award, jointly bestowed by five leading engineering organizations in recognition of ‘great, unselfish, non-technical services by engineers to humanity.’” [4]
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Taylor, Frederick. "The Principles of Scientific Management." New York and London: Harper & Brothers Publishers. 1911. ↵
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"
The Tale of Taylor and Gilbreth.
" All About Lean. November 10, 2013. Accessed July 29, 2019. ↵
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Paules-Bronet, Ileana. "
The Movie 'Cheaper by the Dozen' Was Inspired by a Real 1920's Family With 12 Children.
" Little Things. Accessed July 29, 2019. ↵
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Giges, Nancy. "
Lillian Moller Gilbreth.
" ASME.org. April 30, 2012. Accessed July 29, 2019. ↵
Contributors and Attributions
- Frederick Winslow Taylor. Authored by : Unknown. Located at : https://commons.wikimedia.org/wiki/File:Frederick_Winslow_Taylor_crop.jpg . License : Public Domain: No Known Copyright
- Lillian Moller Gilbreth. Authored by : Harris & Ewing. Located at : https://commons.wikimedia.org/wiki/File:Lillian_Moller_Gilbreth.jpg . License : Public Domain: No Known Copyright


8.3: Skepticism and Cynicism
Cynicism, as mentioned previously, is a major negative component of the police subculture. Cynicism should not be confused with skepticism as they are vastly different. Skepticism refers to critical thinking, which is what we should all strive for. It allows us to question commonly held beliefs that may not be true. It forms the basis of scientific inquiry that has enabled humanity to evolve from a reliance on superstition to an ability to address difficult questions and problems in society using research and reasoned approaches. Skepticism is especially important in law enforcement as it is a critical component of being an objective investigator, one who is willing to view issues from a neutral perspective. According to Kurtz (2010, p.13) “a skeptic is one who is willing to question any claim to truth, asking for clarity in definition, consistency in logic, and adequacy of evidence.” The ability to question any claims of truth is critical for all law enforcement officers who are confronted on a daily basis with people who are disinclined to tell the truth in order to protect their own interests. Furthermore, investigators should strive for empirical evidence that eliminates a biased conclusion that can potentially lead to wrongful accusations and convictions.
Skepticism is endangered when we have difficulty distinguishing between “questioning truth” and “questioning and rejecting truth. Kurtz (2010) defines this as “nihilistic skepticism.” In essence, this is the assertion that nothing should be believed as truth, because it is unprovable. As such, nihilistic skepticism is based upon subjectivity. Nihilistic skepticism is essentially skepticism in which there is no basis for “objective moral judgment” (Kurtz, 2010, p.15).
Cynicism is similar to nihilistic skepticism; however, it embodies an element of pessimism toward everyday occurrences that may ignore the objective truth. Cynicism can include a negative response to morality that illustrates a contempt for community standards. Truth is unimportant to a cynic, and the distrust that is exhibited by the cynic belies the facts. Specifically, law enforcement officers often see themselves in an “us versus them” environment in which officers are attacked from all sides of society. Officers routinely say that the biggest stress they face is not on the streets from dangerous people but in the office before they make it to the streets. This notion is, to a degree, cynical. The objective truth is that the streets are not as safe as the office; however, officers ignore this due to their frustration in dealing with management.
Another example of cynicism is the belief some officers have that the only way to remove sex-trade workers from the area they patrol is to verbally abuse them to the point they feel threatened. In cynical officers’ minds, such behaviour is acceptable even though it may be contrary to community standards and their agency’s values.
Sunahara (2002) draws a further link between cynicism, alienation, and estrangement from management and departmental policy, suggesting that police officers who become alienated are likely to become disengaged from performing their duties and could slip into unethical behaviour. The cause of alienation is unclear, raising the possibility that cynicism begets alienation, or that alienation begets cynicism.
Graves (1996, as cited in Sunahara, 2002, p.12) describes police cynicism as “an attitude of ‘contemptuous distrust of human nature and motives.’  The feeling of alienation stems from this attitude as officers shield themselves from those they believe they cannot trust and from those who are potentially dangerous to them and their families. This feeling is likely the result of working in a negative environment in which officers see and do things that mainstream society is not privy to. Police officers often talk about happenings on the job that “you couldn’t write in a movie script, because they are so unbelievable.” Much of what police officers see is rarely talked about in mainstream society because it is alarming and disturbing. Yet officers are confronted with these issues not only on a daily basis but multiple times a day, depending on where the officer works. It is little wonder that officers in this environment tend to want to alienate themselves from society, and little wonder why alienation, at the very least, isn’t coupled with cynicism.