6.5: The Sample Variance
Descriptive Theory
Recall the basic model of statistics: we have a population of objects of interest, and we have various measurements (variables) that we make on these objects. We select objects from the population and record the variables for the objects in the sample; these become our data. Once again, our first discussion is from a descriptive point of view. That is, we do not assume that the data are generated by an underlying probability distribution. Remember however, that the data themselves form a probability distribution.
Variance and Standard Deviation
Suppose that \(\bs{x} = (x_1, x_2, \ldots, x_n)\) is a sample of size \(n\) from a real-valued variable \(x\). Recall that the sample mean is \[ m = \frac{1}{n} \sum_{i=1}^n x_i \] and is the most important measure of the center of the data set. The sample variance is defined to be \[ s^2 = \frac{1}{n - 1} \sum_{i=1}^n (x_i - m)^2 \] If we need to indicate the dependence on the data vector \(\bs{x}\), we write \(s^2(\bs{x})\). The difference \(x_i - m\) is the deviation of \(x_i\) from the mean \(m\) of the data set. Thus, the variance is the mean square deviation and is a measure of the spread of the data set with respet to the mean. The reason for dividing by \(n - 1\) rather than \(n\) is best understood in terms of the inferential point of view that we discuss in the next section; this definition makes the sample variance an unbiased estimator of the distribution variance. However, the reason for the averaging can also be understood in terms of a related concept.
\(\sum_{i=1}^n (x_i - m) = 0\).
Proof
\(\sum_{i=1}^n (x_i - m) = \sum_{i=1}^n x_i - \sum_{i=1}^n m = n m - n m = 0\).
Thus, if we know \(n - 1\) of the deviations, we can compute the last one. This means that there are only \(n - 1\) freely varying deviations, that is to say, \(n - 1\) degrees of freedom in the set of deviations. In the definition of sample variance, we average the squared deviations, not by dividing by the number of terms, but rather by dividing by the number of degrees of freedom in those terms. However, this argument notwithstanding, it would be reasonable, from a purely descriptive point of view , to divide by \(n\) in the definition of the sample variance. Moreover, when \(n\) is sufficiently large, it hardly matters whether we divide by \(n\) or by \(n - 1\).
In any event, the square root \(s\) of the sample variance \(s^2\) is the sample standard deviation. It is the root mean square deviation and is also a measure of the spread of the data with respect to the mean. Both measures of spread are important. Variance has nicer mathematical properties, but its physical unit is the square of the unit of \(x\). For example, if the underlying variable \(x\) is the height of a person in inches, the variance is in square inches. On the other hand, the standard deviation has the same physical unit as the original variable, but its mathematical properties are not as nice.
Recall that the data set \(\bs{x}\) naturally gives rise to a probability distribution, namely the empirical distribution that places probability \(\frac{1}{n}\) at \(x_i\) for each \(i\). Thus, if the data are distinct, this is the uniform distribution on \(\{x_1, x_2, \ldots, x_n\}\). The sample mean \(m\) is simply the expected value of the empirical distribution. Similarly, if we were to divide by \(n\) rather than \(n - 1\), the sample variance would be the variance of the empirical distribution. Most of the properties and results this section follow from much more general properties and results for the variance of a probability distribution (although for the most part, we give independent proofs).
Measures of Center and Spread
Measures of center and measures of spread are best thought of together, in the context of an error function . The error function measures how well a single number \(a\) represents the entire data set \(\bs{x}\). The values of \(a\) (if they exist) that minimize the error functions are our measures of center; the minimum value of the error function is the corresponding measure of spread. Of course, we hope for a single value of \(a\) that minimizes the error function, so that we have a unique measure of center.
Let's apply this procedure to the mean square error function defined by \[ \mse(a) = \frac{1}{n - 1} \sum_{i=1}^n (x_i - a)^2, \quad a \in \R \] Minimizing \(\mse\) is a standard problem in calculus.
The graph of \(\mse\) is a parabola opening upward.
- \(\mse\) is minimized when \(a = m\), the sample mean.
- The minimum value of \(\mse\) is \(s^2\), the sample variance.
Proof
We can tell from the form of \(\mse\) that the graph is a parabola opening upward. Taking the derivative gives \[ \frac{d}{da} \mse(a) = -\frac{2}{n - 1}\sum_{i=1}^n (x_i - a) = -\frac{2}{n - 1}(n m - n a) \] Hence \(a = m\) is the unique value that minimizes \(\mse\). Of course, \(\mse(m) = s^2\).
Trivially, if we defined the mean square error function by dividing by \(n\) rather than \(n - 1\), then the minimum value would still occur at \(m\), the sample mean, but the minimum value would be the alternate version of the sample variance in which we divide by \(n\). On the other hand, if we were to use the root mean square deviation function \(\text{rmse}(a) = \sqrt{\mse(a)}\), then because the square root function is strictly increasing on \([0, \infty)\), the minimum value would again occur at \(m\), the sample mean, but the minimum value would be \(s\), the sample standard deviation. The important point is that with all of these error functions, the unique measure of center is the sample mean, and the corresponding measures of spread are the various ones that we are studying.
Next, let's apply our procedure to the mean absolute error function defined by \[ \mae(a) = \frac{1}{n - 1} \sum_{i=1}^n \left|x_i - a\right|, \quad a \in \R \]
The mean absolute error function satisfies the following properties:
- \(\mae\) is a continuous function.
- The graph of \(\mae\) consists of lines.
- The slope of the line at \(a\) depends on where \(a\) is in the data set \(\bs{x}\).
Proof
For parts (a) and (b), note that for each \(i\), \(\left|x_i - a\right|\) is a continuous function of \(a\) with the graph consisting of two lines (of slopes \(\pm 1\)) meeting at \(x_i\).
Mathematically, \(\mae\) has some problems as an error function. First, the function will not be smooth (differentiable) at points where two lines of different slopes meet. More importantly, the values that minimize mae may occupy an entire interval, thus leaving us without a unique measure of center. The error function exercises below will show you that these pathologies can really happen. It turns out that \(\mae\) is minimized at any point in the median interval of the data set \(\bs{x}\). The proof of this result follows from a much more general result for probability distributions. Thus, the medians are the natural measures of center associated with \(\mae\) as a measure of error, in the same way that the sample mean is the measure of center associated with the \(\mse\) as a measure of error.
Properties
In this section, we establish some essential properties of the sample variance and standard deviation. First, the following alternate formula for the sample variance is better for computational purposes, and for certain theoretical purposes as well.
The sample variance can be computed as \[ s^2 = \frac{1}{n - 1} \sum_{i=1}^n x_i^2 - \frac{n}{n - 1} m^2 \]
Proof
Note that \begin{align} \sum_{i=1}^n (x_i - m)^2 & = \sum_{i=1}^n \left(x_i^2 - 2 m x_i + m^2\right) = \sum_{i=1}^n x_i^2 - 2 m \sum_{i=1}^n x_i - \sum_{i=1}^n m\\ & = \sum_{i=1}^n x_i^2 - 2 n m^2 + n m^2 = \sum_{i=1}^n x_i^2 - n m^2 \end{align} Dividing by \(n - 1\) gives the result.
If we let \(\bs{x}^2 = (x_1^2, x_2^2, \ldots, x_n^2)\) denote the sample from the variable \(x^2\), then the computational formula in the last exercise can be written succinctly as \[ s^2(\bs{x}) = \frac{n}{n - 1} \left[m(\bs{x}^2) - m^2(\bs{x})\right] \] The following theorem gives another computational formula for the sample variance, directly in terms of the variables and thus without the computation of an intermediate statistic.
The sample variance can be computed as \[ s^2 = \frac{1}{2 n (n - 1)} \sum_{i=1}^n \sum_{j=1}^n (x_i - x_j)^2 \]
Proof
Note that \begin{align} \frac{1}{2 n} \sum_{i=1}^n \sum_{j=1}^n (x_i - x_j)^2 & = \frac{1}{2 n} \sum_{i=1}^n \sum_{j=1}^n (x_i - m + m - x_j)^2 \\ & = \frac{1}{2 n} \sum_{i=1}^n \sum_{j=1}^n \left[(x_i - m)^2 + 2 (x_i - m)(m - x_j) + (m - x_j)^2\right] \\ & = \frac{1}{2 n} \sum_{i=1}^n \sum_{j=1}^n (x_i - m)^2 + \frac{1}{n} \sum_{i=1}^n \sum_{j=1}^n (x_i - m)(m - x_j) + \frac{1}{2 n} \sum_{i=1}^n \sum_{j=1}^n (m - x_j)^2 \\ & = \frac{1}{2} \sum_{i=1}^n (x_i - m)^2 + 0 + \frac{1}{2} \sum_{j=1}^n (m - x_j)^2 \\ & = \sum_{i=1}^n (x_i - m)^2 \end{align} Dividing by \(n - 1\) gives the result.
The sample variance is nonnegative:
- \(s^2 \ge 0\)
- \(s^2 = 0\) if and only if \(x_i = x_j\) for each \(i, \; j \in \{1, 2, \ldots, n\}\).
Proof
Part (a) is obvious. For part (b) note that if \(s^2 = 0\) then \(x_i = m\) for each \(i\). Conversely, if \(\bs{x}\) is a constant vector, then \(m\) is that same constant.
Thus, \(s^2 = 0\) if and only if the data set is constant (and then, of course, the mean is the common value).
If \(c\) is a constant then
- \(s^2(c \, \bs{x}) = c^2 \, s^2(\bs{x})\)
- \(s(c \, \bs{x}) = \left|c\right| \, s(\bs{x})\)
Proof
For part (a), recall that \(m(c \bs{x}) = c m(\bs{x})\). Hence \[ s^2(c \bs{x}) = \frac{1}{n - 1}\sum_{i=1}^n \left[c x_i - c m(\bs{x})\right]^2 = \frac{1}{n - 1} \sum_{i=1}^n c^2 \left[x_i - m(\bs{x})\right]^2 = c^2 s^2(\bs{x}) \]
If \(\bs{c}\) is a sample of size \(n\) from a constant \(c\) then
- \(s^2(\bs{x} + \bs{c}) = s^2(\bs{x})\).
- \(s(\bs{x} + \bs{c}) = s(\bs{x})\)
Proof
Recall that \(m(\bs{x} + \bs{c}) = m(\bs{x}) + c\). Hence \[ s^2(\bs{x} + \bs{c}) = \frac{1}{n - 1} \sum_{i=1}^n \left\{(x_i + c) - \left[m(\bs{x}) + c\right]\right\}^2 = \frac{1}{n - 1} \sum_{i=1}^n \left[x_i - m(\bs{x})\right]^2 = s^2(\bs{x})\]
As a special case of these results, suppose that \(\bs{x} = (x_1, x_2, \ldots, x_n)\) is a sample of size \(n\) corresponding to a real variable \(x\), and that \(a\) and \(b\) are constants. The sample corresponding to the variable \(y = a + b x\), in our vector notation, is \(\bs{a} + b \bs{x}\). Then \(m(\bs{a} + b \bs{x}) = a + b m(\bs{x})\) and \(s(\bs{a} + b \bs{x}) = \left|b\right| s(\bs{x})\). Linear transformations of this type, when \(b \gt 0\), arise frequently when physical units are changed. In this case, the transformation is often called a location-scale transformation; \(a\) is the location parameter and \(b\) is the scale parameter. For example, if \(x\) is the length of an object in inches, then \(y = 2.54 x\) is the length of the object in centimeters. If \(x\) is the temperature of an object in degrees Fahrenheit, then \(y = \frac{5}{9}(x - 32)\) is the temperature of the object in degree Celsius.
Now, for \(i \in \{1, 2, \ldots, n\}\), let \( z_i = (x_i - m) / s\). The number \(z_i\) is the standard score associated with \(x_i\). Note that since \(x_i\), \(m\), and \(s\) have the same physical units, the standard score \(z_i\) is dimensionless (that is, has no physical units); it measures the directed distance from the mean \(m\) to the data value \(x_i\) in standard deviations.
The sample of standard scores \(\bs{z} = (z_1, z_2, \ldots, z_n)\) has mean 0 and variance 1. That is,
- \(m(\bs{z}) = 0\)
- \(s^2(\bs{z}) = 1\)
Proof
These results follow from Theroems 7 and 8. In vector notation, note that \(\bs{z} = (\bs{x} - \bs{m})/s\). Hence \(m(\bs{z}) = (m - m) / s = 0\) and \(s(\bs{z}) = s / s = 1\).
Approximating the Variance
Suppose that instead of the actual data \(\bs{x}\), we have a frequency distribution corresponding to a partition with classes (intervals) \((A_1, A_2, \ldots, A_k)\), class marks (midpoints of the intervals) \((t_1, t_2, \ldots, t_k)\), and frequencies \((n_1, n_2, \ldots, n_k)\). Recall that the relative frequency of class \(A_j\) is \(p_j = n_j / n\). In this case, approximate values of the sample mean and variance are, respectively,
\begin{align} m & = \frac{1}{n} \sum_{j=1}^k n_j \, t_j = \sum_{j = 1}^k p_j \, t_j \\ s^2 & = \frac{1}{n - 1} \sum_{j=1}^k n_j (t_j - m)^2 = \frac{n}{n - 1} \sum_{j=1}^k p_j (t_j - m)^2 \end{align}
These approximations are based on the hope that the data values in each class are well represented by the class mark. In fact, these are the standard definitions of sample mean and variance for the data set in which \(t_j\) occurs \(n_j\) times for each \(j\).
Inferential Statistics
We continue our discussion of the sample variance, but now we assume that the variables are random. Thus, suppose that we have a basic random experiment, and that \(X\) is a real-valued random variable for the experiment with mean \(\mu\) and standard deviation \(\sigma\). We will need some higher order moments as well. Let \(\sigma_3 = \E\left[(X - \mu)^3\right]\) and \(\sigma_4 = \E\left[(X - \mu)^4\right]\) denote the 3rd and 4th moments about the mean. Recall that \(\sigma_3 \big/ \sigma^3 = \skw(X)\), the skewness of \(X\), and \(\sigma_4 \big/ \sigma^4 = \kur(X)\), the kurtosis of \(X\). We assume that \(\sigma_4 \lt \infty\).
We repeat the basic experiment \(n\) times to form a new, compound experiment, with a sequence of independent random variables \(\bs{X} = (X_1, X_2, \ldots, X_n)\), each with the same distribution as \(X\). In statistical terms, \(\bs{X}\) is a random sample of size \(n\) from the distribution of \(X\). All of the statistics above make sense for \(\bs{X}\), of course, but now these statistics are random variables. We will use the same notationt, except for the usual convention of denoting random variables by capital letters. Finally, note that the deterministic properties and relations established above still hold.
In addition to being a measure of the center of the data \(\bs{X}\), the sample mean \[ M = \frac{1}{n} \sum_{i=1}^n X_i \] is a natural estimator of the distribution mean \(\mu\). In this section, we will derive statistics that are natural estimators of the distribution variance \(\sigma^2\). The statistics that we will derive are different, depending on whether \(\mu\) is known or unknown; for this reason, \(\mu\) is referred to as a nuisance parameter for the problem of estimating \(\sigma^2\).
A Special Sample Variance
First we will assume that \(\mu\) is known. Although this is almost always an artificial assumption, it is a nice place to start because the analysis is relatively easy and will give us insight for the standard case. A natural estimator of \(\sigma^2\) is the following statistic, which we will refer to as the special sample variance . \[ W^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2 \]
\(W^2\) is the sample mean for a random sample of size \(n\) from the distribution of \((X - \mu)^2\), and satisfies the following properties:
- \(\E\left(W^2\right) = \sigma^2\)
- \(\var\left(W^2\right) = \frac{1}{n}\left(\sigma_4 - \sigma^4\right)\)
- \(W^2 \to \sigma^2\) as \(n \to \infty\) with probability 1
- The distribution of \(\sqrt{n}\left(W^2 - \sigma^2\right) \big/ \sqrt{\sigma_4 - \sigma^4}\) converges to the standard normal distribution as \(n \to \infty\).
Proof
These result follow immediately from standard results in the section on the Law of Large Numbers and the section on the Central Limit Theorem. For part (b), note that \[\var\left[(X - \mu)^2\right] = \E\left[(X - \mu)^4\right] -\left(\E\left[(X - \mu)^2\right]\right)^2 = \sigma_4 - \sigma^4\]
In particular part (a) means that \(W^2\) is an unbiased estimator of \(\sigma^2\). From part (b), note that \(\var(W^2) \to 0\) as \(n \to \infty\); this means that \(W^2\) is a consistent estimator of \(\sigma^2\). The square root of the special sample variance is a special version of the sample standard deviation , denoted \(W\).
\(\E(W) \le \sigma\). Thus, \(W\) is a negativley biased estimator that tends to underestimate \(\sigma\).
Proof
This follows from the unbiased property and Jensen's inequality. Since \(w \mapsto \sqrt{w}\) is concave downward on \([0, \infty)\), we have \(\E(W) = \E\left(\sqrt{W^2}\right) \le \sqrt{\E\left(W^2\right)} = \sqrt{\sigma^2} = \sigma\).
Next we compute the covariance and correlation between the sample mean and the special sample variance.
The covariance and correlation of \(M\) and \(W^2\) are
- \(\cov\left(M, W^2\right) = \sigma_3 / n\).
- \(\cor\left(M, W^2\right) = \sigma^3 \big/ \sqrt{\sigma^2 (\sigma_4 - \sigma^4)}\)
Proof
- From the bilinearity of the covariance operator and by independence, \[ \cov\left(M, W^2\right) = \cov\left[\frac{1}{n}\sum_{i=1}^n X_i, \frac{1}{n} \sum_{j=1}^n (X_j - \mu)^2\right] = \frac{1}{n^2} \sum_{i=1}^n \cov\left[X_i, (X_i - \mu)^2\right] \] But \(\cov\left[X_i, (X_i - \mu)^2\right] = \cov\left[X_i - \mu, (X_i - \mu)^2\right] = \E\left[(X_i - \mu)^3\right] - \E(X_i - \mu) \E\left[(X_i - \mu)^2\right] = \sigma_3\). Substituting gives the result.
- This follows from part (a), the unbiased property, and our previous result that \(\var(M) = \sigma^2 / n\).
Note that the correlation does not depend on the sample size, and that the sample mean and the special sample variance are uncorrelated if \(\sigma_3 = 0\) (equivalently \(\skw(X) = 0\)).
The Standard Sample Variance
Consider now the more realistic case in which \(\mu\) is unknown. In this case, a natural approach is to average, in some sense, the squared deviations \((X_i - M)^2\) over \(i \in \{1, 2, \ldots, n\}\). It might seem that we should average by dividing by \(n\). However, another approach is to divide by whatever constant would give us an unbiased estimator of \(\sigma^2\). This constant turns out to be \(n - 1\), leading to the standard sample variance : \[ S^2 = \frac{1}{n - 1} \sum_{i=1}^n (X_i - M)^2 \]
\(\E\left(S^2\right) = \sigma^2\).
Proof
By expanding (as was shown in the last section), \[ \sum_{i=1}^n (X_i - M)^2 = \sum_{i=1}^n X_i^2 - n M^2 \] Recall that \(\E(M) = \mu\) and \(\var(M) = \sigma^2 / n\). Taking expected values in the displayed equation gives \[ \E\left(\sum_{i=1}^n (X_i - M)^2\right) = \sum_{i=1}^n (\sigma^2 + \mu^2) - n \left(\frac{\sigma^2}{n} + \mu^2\right) = n (\sigma^2 + \mu^2) -n \left(\frac{\sigma^2}{n} + \mu^2\right) = (n - 1) \sigma^2 \]
Of course, the square root of the sample variance is the sample standard deviation , denoted \(S\).
\(\E(S) \le \sigma\). Thus, \(S\) is a negativley biased estimator than tends to underestimate \(\sigma\).
Proof
The proof is exactly the same as for the special standard variance .
\(S^2 \to \sigma^2\) as \(n \to \infty\) with probability 1.
Proof
This follows from the strong law of large numbers. Recall again that \[ S^2 = \frac{1}{n - 1} \sum_{i=1}^n X_i^2 - \frac{n}{n - 1} M^2 = \frac{n}{n - 1}[M(\bs{X}^2) - M^2(\bs{X})] \] But with probability 1, \(M(\bs{X}^2) \to \sigma^2 + \mu^2\) as \(n \to \infty\) and \(M^2(\bs{X}) \to \mu^2\) as \(n \to \infty\).
Since \(S^2\) is an unbiased estimator of \(\sigma^2\), the variance of \(S^2\) is the mean square error , a measure of the quality of the estimator.
\(\var\left(S^2\right) = \frac{1}{n} \left( \sigma_4 - \frac{n - 3}{n - 1} \sigma^4 \right)\).
Proof
Recall from the result above that \[ S^2 = \frac{1}{2 n (n - 1)} \sum_{i=1}^n \sum_{j=1}^n (X_i - X_j)^2 \] Hence, using the bilinear property of covariance we have \[ \var(S^2) = \cov(S^2, S^2) = \frac{1}{4 n^2 (n - 1)^2} \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \sum_{k=1}^n \cov[(X_i - X_j)^2, (X_k - X_l)^2] \] We compute the covariances in this sum by considering disjoint cases:
- \(\cov\left[(X_i - X_j)^2, (X_k - X_l)^2\right] = 0\) if \(i = j\) or \(k = l\), and there are \(2 n^3 - n^2\) such terms.
- \(\cov\left[(X_i - X_j)^2, (X_k - X_l)^2\right] = 0\) if \(i, j, k, l\) are distinct, and there are \(n (n - 1)(n - 2) (n - 3)\) such terms.
- \(\cov\left[(X_i - X_j)^2, (X_k - X_l)^2\right] = 2 \sigma_4 + 2 \sigma^4\) if \(i \ne j\) and \(\{k, l\} = \{i, j\}\), and there are \(2 n (n - 1)\) such terms.
- \(\cov\left[(X_i - X_j)^2, (X_k - X_l)^2\right] = \sigma_4 - \sigma^4\) if \(i \ne j\), \(k \ne l\) and \(\#(\{i, j\} \cap \{k, l\}) = 1\), and there are \(4 n (n - 1)(n - 2)\) such terms.
Substituting gives the result.
Note that \(\var(S^2) \to 0\) as \(n \to \infty\), and hence \(S^2\) is a consistent estimator of \(\sigma^2\). On the other hand, it's not surprising that the variance of the standard sample variance (where we assume that \(\mu\) is unknown) is greater than the variance of the special standard variance (in which we assume \(\mu\) is known).
\(\var\left(S^2\right) \gt \var\left(W^2\right)\).
Proof
From the formula above for the variance of \( W^2 \), the previous result for the variance of \( S^2 \), and simple algebra, \[ \var\left(S^2\right) - \var\left(W^2\right) = \frac{2}{n (n - 1)} \sigma^4 \] Note however that the difference goes to 0 as \(n \to \infty\).
Next we compute the covariance between the sample mean and the sample variance.
The covariance and correlation between the sample mean and sample variance are
- \(\cov\left(M, S^2\right) = \sigma_3 / n\)
- \(\cor\left(M, S^2\right) = \frac{\sigma_3}{\sigma \sqrt{\sigma_4 - \sigma^4 (n - 3) / (n - 1)}}\)
Proof
-
Recall again that \[ M = \frac{1}{n} \sum_{i=1}^n X_i, \quad S^2 = \frac{1}{2 n (n - 1)} \sum_{j=1}^n \sum_{k=1}^n (X_j - X_k)^2 \] Hence, using the bilinear property of covariance we have \[ \cov(M, S^2) = \frac{1}{2 n^2 (n - 1)} \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \cov[X_i, (X_j - X_k)^2] \] We compute the covariances in this sum by considering disjoint cases:
- \(\cov\left[X_i, (X_j - X_k)^2\right] = 0\) if \(j = k\), and there are \(n^2\) such terms.
- \(\cov\left[X_i, (X_j - X_k)^2\right] = 0\) if \(i, j, k\) are distinct, and there are \(n (n - 1)(n - 2)\) such terms.
- \(\cov\left[X_i, (X_j - X_k)^2\right] = \sigma_3\) if \(j \ne k\) and \(i \in \{j, k\}\), and there are \(2 n (n - 1)\) such terms.
- This follows follows from part(a), the result above on the variance of \( S^2 \), and \(\var(M) = \sigma^2 / n\).
In particular, note that \(\cov(M, S^2) = \cov(M, W^2)\). Again, the sample mean and variance are uncorrelated if \(\sigma_3 = 0\) so that \(\skw(X) = 0\). Our last result gives the covariance and correlation between the special sample variance and the standard one. Curiously, the covariance the same as the variance of the special sample variance.
The covariance and correlation between \(W^2\) and \(S^2\) are
- \(\cov\left(W^2, S^2\right) = (\sigma_4 - \sigma^4) / n\)
- \(\cor\left(W^2, S^2\right) = \sqrt{\frac{\sigma_4 - \sigma^4}{\sigma_4 - \sigma^4 (n - 3) / (n - 1)}}\)
Proof
-
Recall again that \[ W^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2, \quad S^2 = \frac{1}{2 n (n - 1)} \sum_{j=1}^n \sum_{k=1}^n (X_j - X_k)^2\] so by the bilinear property of covariance we have \[ \cov(W^2, S^2) = \frac{1}{2 n^2 (n - 1)} \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n \cov[(X_i - \mu)^2, (X_j - X_k)^2] \] Once again, we compute the covariances in this sum by considering disjoint cases:
- \(\cov[(X_i - \mu)^2, (X_j - X_k)^2] = 0\) if \(j = k\), and there are \(n^2\) such terms.
- \(\cov[(X_i - \mu)^2, (X_j - X_k)^2] = 0\) if \(i, j, k\) are distinct, and there are \(n (n - 1)(n - 2)\) such terms.
- \(\cov[(X_i - \mu)^2, (X_j - X_k)^2] = \sigma_4 - \sigma^4\) if \(j \ne k\) and \(i \in \{j, k\}\), and there are \(2 n (n - 1)\) such terms.
- This follows from part (a) and the formulas above for the variance of \( W^2 \) and the variance of \( V^2 \)
Note that \(\cor\left(W^2, S^2\right) \to 1\) as \(n \to \infty\), not surprising since with probability 1, \(S^2 \to \sigma^2\) and \(W^2 \to \sigma^2\) as \(n \to \infty\).
A particularly important special case occurs when the sampling distribution is normal. This case is explored in the section on Special Properties of Normal Samples.
Exercises
Basic Properties
Suppose that \(x\) is the temperature (in degrees Fahrenheit) for a certain type of electronic component after 10 hours of operation. A sample of 30 components has mean 113° and standard deviation \(18°\).
- Classify \(x\) by type and level of measurement.
- Find the sample mean and standard deviation if the temperature is converted to degrees Celsius. The transformation is \(y = \frac{5}{9}(x - 32)\).
Answer
- continuous, interval
- \(m = 45°\), \(s = 10°\)
Suppose that \(x\) is the length (in inches) of a machined part in a manufacturing process. A sample of 50 parts has mean 10.0 and standard deviation 2.0.
- Classify \(x\) by type and level of measurement.
- Find the sample mean if length is measured in centimeters. The transformation is \(y = 2.54 x\).
Answer
- continuous, ratio
- \(m = 25.4\), \(s = 5.08\)
Professor Moriarity has a class of 25 students in her section of Stat 101 at Enormous State University (ESU). The mean grade on the first midterm exam was 64 (out of a possible 100 points) and the standard deviation was 16. Professor Moriarity thinks the grades are a bit low and is considering various transformations for increasing the grades. In each case below give the mean and standard deviation of the transformed grades, or state that there is not enough information.
- Add 10 points to each grade, so the transformation is \(y = x + 10\).
- Multiply each grade by 1.2, so the transformation is \(z = 1.2 x\)
- Use the transformation \(w = 10 \sqrt{x}\). Note that this is a non-linear transformation that curves the grades greatly at the low end and very little at the high end. For example, a grade of 100 is still 100, but a grade of 36 is transformed to 60.
One of the students did not study at all, and received a 10 on the midterm. Professor Moriarity considers this score to be an outlier.
- Find the mean and standard deviation if this score is omitted.
Answer
- \(m = 74\), \(s = 16\)
- \(m = 76.8\), \(s = 19.2\)
- Not enough information
- \(m = 66.25\), \(s = 11.62\)
Computational Exercises
All statistical software packages will compute means, variances and standard deviations, draw dotplots and histograms, and in general perform the numerical and graphical procedures discussed in this section. For real statistical experiments, particularly those with large data sets, the use of statistical software is essential. On the other hand, there is some value in performing the computations by hand, with small, artificial data sets, in order to master the concepts and definitions. In this subsection, do the computations and draw the graphs with minimal technological aids.
Suppose that \(x\) is the number of math courses completed by an ESU student. A sample of 10 ESU students gives the data \(\bs{x} = (3, 1, 2, 0, 2, 4, 3, 2, 1, 2)\).
- Classify \(x\) by type and level of measurement.
- Sketch the dotplot.
- Construct a table with rows corresponding to cases and columns corresponding to \(i\), \(x_i\), \(x_i - m\), and \((x_i - m)^2\). Add rows at the bottom in the \(i\) column for totals and means.
Answer
- discrete, ratio
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\(i\) \(x_i\) \(x_i - m\) \((x_i - m)^2\) \(1\) \(3\) \(1\) \(1\) \(2\) \(1\) \(-1\) \(1\) \(3\) \(2\) \(0\) \(0\) \(4\) \(0\) \(-2\) \(4\) \(5\) \(2\) \(0\) \(0\) \(6\) \(4\) \(2\) \(4\) \(7\) \(3\) \(1\) \(1\) \(8\) \(2\) \(0\) \(0\) \(9\) \(1\) \(-1\) \(1\) \(10\) \(2\) \(0\) \(0\) Total 20 0 14 Mean 2 0 \(14/9\)
Suppose that a sample of size 12 from a discrete variable \(x\) has empirical density function given by \(f(-2) = 1/12\), \(f(-1) = 1/4\), \(f(0) = 1/3\), \(f(1) = 1/6\), \(f(2) = 1/6\).
- Sketch the graph of \(f\).
- Compute the sample mean and variance.
- Give the sample values, ordered from smallest to largest.
Answer
- \( m = 1/12\), \(s^2 = 203/121\)
- \((-2, -1, -1, -1, 0, 0, 0, 0, 1, 1, 2, 2)\)
The following table gives a frequency distribution for the commuting distance to the math/stat building (in miles) for a sample of ESU students.
| Class | Freq | Rel Freq | Density | Cum Freq | Cum Rel Freq | Midpoint |
|---|---|---|---|---|---|---|
| \((0, 2]\) | 6 | |||||
| \((2, 6]\) | 16 | |||||
| \((6, 10]\) | 18 | |||||
| \((10, 20])\) | 10 | |||||
| Total |
- Complete the table
- Sketch the density histogram
- Sketch the cumulative relative frquency ogive.
- Compute an approximation to the mean and standard deviation.
Answer
-
Class Freq Rel Freq Density Cum Freq Cum Rel Freq Midpoint \((0, 2]\) 6 0.12 0.06 6 0.12 1 \((2, 6]\) 16 0.32 0.08 22 0.44 4 \((6, 10]\) 18 0.36 0.09 40 0.80 8 \((10, 20]\) 10 0.20 0.02 50 1 15 Total 50 1 - \(m = 7.28\), \(s = 4.549\)
Error Function Exercises
In the error function app, select root mean square error. As you add points, note the shape of the graph of the error function, the value that minimizes the function, and the minimum value of the function.
In the error function app, select mean absolute error. As you add points, note the shape of the graph of the error function, the values that minimizes the function, and the minimum value of the function.
Suppose that our data vector is \((2, 1, 5, 7)\). Explicitly give \(\mae\) as a piecewise function and sketch its graph. Note that
- All values of \(a \in [2, 5]\) minimize \(\mae\).
- \(\mae\) is not differentiable at \(a \in \{1, 2, 5, 7\}\).
Suppose that our data vector is \((3, 5, 1)\). Explicitly give \(\mae\) as a piecewise function and sketch its graph. Note that
- \(\mae\) is minimized at \(a = 3\).
- \(\mae\) is not differentiable at \(a \in \{1, 3, 5\}\).
Simulation Exercises
Many of the apps in this project are simulations of experiments with a basic random variable of interest. When you run the simulation, you are performing independent replications of the experiment. In most cases, the app displays the standard deviation of the distribution, both numerically in a table and graphically as the radius of the blue, horizontal bar in the graph box. When you run the simulation, the sample standard deviation is also displayed numerically in the table and graphically as the radius of the red horizontal bar in the graph box.
In the binomial coin experiment, the random variable is the number of heads. For various values of the parameters \(n\) (the number of coins) and \(p\) (the probability of heads), run the simulation 1000 times and compare the sample standard deviation to the distribution standard deviation.
In the simulation of the matching experiment, the random variable is the number of matches. For selected values of \(n\) (the number of balls), run the simulation 1000 times and compare the sample standard deviation to the distribution standard deviation.
Run the simulation of the gamma experiment 1000 times for various values of the rate parameter \(r\) and the shape parameter \(k\). Compare the sample standard deviation to the distribution standard deviation.
Probability Exercises
Suppose that \(X\) has probability density function \(f(x) = 12 \, x^2 \, (1 - x)\) for \(0 \le x \le 1\). The distribution of \(X\) is a member of the beta family. Compute each of the following
- \(\mu = \E(X)\)
- \(\sigma^2 = \var(X)\)
- \(d_3 = \E\left[(X - \mu)^3\right]\)
- \(d_4 = \E\left[(X - \mu)^4\right]\)
Answer
- \(3/5\)
- \(1/25\)
- \(-2/875\)\)
- \(33/8750\)
Suppose now that \((X_1, X_2, \ldots, X_{10})\) is a random sample of size 10 from the beta distribution in the previous problem. Find each of the following:
- \(\E(M)\)
- \(\var(M)\)
- \(\E\left(W^2\right)\)
- \(\var\left(W^2\right)\)
- \(\E\left(S^2\right)\)
- \(\var\left(S^2\right)\)
- \(\cov\left(M, W^2\right)\)
- \(\cov\left(M, S^2\right)\)
- \(\cov\left(W^2, S^2\right)\)
Answer
- \(3/5\)
- \(1/250\)
- \(1/25\)
- \(19/87\,500\)
- \(1/25\)
- \(199/787\,500\)
- \(-2/8750\)
- \(-2/8750\)
- \(19/87\,500\)
Suppose that \(X\) has probability density function \(f(x) = \lambda e^{-\lambda x}\) for \(0 \le x \lt \infty\), where \(\lambda \gt 0\) is a parameter. Thus \(X\) has the exponential distribution with rate parameter \(\lambda\). Compute each of the following
- \(\mu = \E(X)\)
- \(\sigma^2 = \var(X)\)
- \(d_3 = \E\left[(X - \mu)^3\right]\)
- \(d_4 = \E\left[(X - \mu)^4\right]\)
Answer
- \(1/\lambda\)
- \(1/\lambda^2\)
- \(2/\lambda^3\)
- \(9/\lambda^4\)
Suppose now that \((X_1, X_2, \ldots, X_5)\) is a random sample of size 5 from the exponential distribution in the previous problem. Find each of the following:
- \(\E(M)\)
- \(\var(M)\)
- \(\E\left(W^2\right)\)
- \(\var\left(W^2\right)\)
- \(\E\left(S^2\right)\)
- \(\var\left(S^2\right)\)
- \(\cov\left(M, W^2\right)\)
- \(\cov\left(M, S^2\right)\)
- \(\cov\left(W^2, S^2\right)\)
Answer
- \(1/\lambda\)
- \(1/5 \lambda^2\)
- \(1/\lambda^2\)
- \(8/5 \lambda^4\)
- \(1/\lambda^2\)
- \(17/10 \lambda^4\)
- \(2/5 \lambda^3\)
- \(2/5 \lambda^3\)
- \(8/5 \lambda^4\)
Recall that for an ace-six flat die , faces 1 and 6 have probability \(\frac{1}{4}\) each, while faces 2, 3, 4, and 5 have probability \(\frac{1}{8}\) each. Let \(X\) denote the score when an ace-six flat die is thrown. Compute each of the following:
- \(\mu = \E(X)\)
- \(\sigma^2 = \var(X)\)
- \(d_3 = \E\left[(X - \mu)^3\right]\)
- \(d_4 = \E\left[(X - \mu)^4\right]\)
Answer
- \(7/2\)
- \(15/4\)
- \(0\)
- \(333/16\)
Suppose now that an ace-six flat die is tossed 8 times. Find each of the following:
- \(\E(M)\)
- \(\var(M)\)
- \(\E\left(W^2\right)\)
- \(\var\left(W^2\right)\)
- \(\E\left(S^2\right)\)
- \(\var\left(S^2\right)\)
- \(\cov\left(M, W^2\right)\)
- \(\cov\left(M, S^2\right)\)
- \(\cov\left(W^2, S^2\right)\)
Answer
- \(7/2\)
- \(15/32\)
- \(15/4\)
- \(27/32\)
- \(15/4\)
- \(207/512\)
- \(0\)
- \(0\)
- \(27/32\)
Data Analysis Exercises
Statistical software should be used for the problems in this subsection.
Consider the petal length and species variables in Fisher's iris data.
- Classify the variables by type and level of measurement.
- Compute the sample mean and standard deviation, and plot a density histogram for petal length.
- Compute the sample mean and standard deviation, and plot a density histogram for petal length by species.
Answers
- petal length: continuous, ratio. species: discrete, nominal
- \(m = 37.8\), \(s = 17.8\)
- \(m(0) = 14.6\), \(s(0) = 1.7\); \(m(1) = 55.5\), \(s(1) = 30.5\); \(m(2) = 43.2\), \(s(2) = 28.7\)
Consider the erosion variable in the Challenger data set.
- Classify the variable by type and level of measurement.
- Compute the mean and standard deviation
- Plot a density histogram with the classes \([0, 5)\), \([5, 40)\), \([40, 50)\), \([50, 60)\).
Answer
- continuous, ratio
- \(m = 7.7\), \(s = 17.2\)
Consider Michelson's velocity of light data.
- Classify the variable by type and level of measurement.
- Plot a density histogram.
- Compute the sample mean and standard deviation.
- Find the sample mean and standard deviation if the variable is converted to \(\text{km}/\text{hr}\). The transformation is \(y = x + 299\,000\)
Answer
- continuous, interval
- \(m = 852.4\), \(s = 79.0\)
- \(m = 299\,852.4\), \(s = 79.0\)
Consider Short's paralax of the sun data.
- Classify the variable by type and level of measurement.
- Plot a density histogram.
- Compute the sample mean and standard deviation.
- Find the sample mean and standard deviation if the variable is converted to degrees. There are 3600 seconds in a degree.
- Find the sample mean and standard deviation if the variable is converted to radians. There are \(\pi/180\) radians in a degree.
Answer
- continuous, ratio
- \(m = 8.616\), \(s = 0.749\)
- \(m = 0.00239\), \(s = 0.000208\)
- \(m = 0.0000418\), \(s = 0.00000363\)
Consider Cavendish's density of the earth data.
- Classify the variable by type and level of measurement.
- Compute the sample mean and standard deviation.
- Plot a density histogram.
Answer
- continuous, ratio
- \(m = 5.448\), \(s = 0.221\)
Consider the M&M data.
- Classify the variables by type and level of measurement.
- Compute the sample mean and standard deviation for each color count variable.
- Compute the sample mean and standard deviation for the total number of candies.
- Plot a relative frequency histogram for the total number of candies.
- Compute the sample mean and standard deviation, and plot a density histogram for the net weight.
Answer
- color counts: discrete ratio. net weight: continuous ratio.
- \(m(r) = 9.60\), \(s(r) = 4.12\); \(m(g) = 7.40\), \(s(g) = 0.57\); \(m(bl) = 7.23\), \(s(bl) = 4.35\); \(m(o) = 6.63\), \(s(0) = 3.69\); \(m(y) = 13.77\), \(s(y) = 6.06\); \(m(br) = 12.47\), \(s(br) = 5.13\)
- \(m(n) = 57.10\), \(s(n) = 2.4\)
- \(m(w) = 49.215\), \(s(w) = 1.522\)
Consider the body weight, species, and gender variables in the Cicada data.
- Classify the variables by type and level of measurement.
- Compute the relative frequency function for species and plot the graph.
- Compute the relative frequeny function for gender and plot the graph.
- Compute the sample mean and standard deviation, and plot a density histogram for body weight.
- Compute the sample mean and standard deviation, and plot a density histogrm for body weight by species.
- Compute the sample mean and standard deviation, and plot a density histogram for body weight by gender.
Answer
- body weight: continuous, ratio. species: discrete, nominal. gender: discrete, nominal.
- \(f(0) = 0.423\), \(f(1) = 0.519\), \(f(2) = 0.058\)
- \(f(0) = 0.567\), \(f(1) = 0.433\)
- \(m = 0.180\), \(s = 0.059\)
- \(m(0) = 0.168\), \(s(0) = 0.054\); \(m(1) = 0.185\), \(s(1) = 0.185\); \(m(2) = 0.225\), \(s(2) = 0.107\)
- \(m(0) = 0.206\), \(s(0) = 0.052\); \(m(1) = 0.145\), \(s(1) = 0.051\)
Consider Pearson's height data.
- Classify the variables by type and level of measurement.
- Compute the sample mean and standard deviation, and plot a density histogram for the height of the father.
- Compute the sample mean and standard deviation, and plot a density histogram for the height of the son.
Answer
- continuous ratio
- \(m(x) = 67.69\), \(s(x) = 2.75\)
- \(m(y) = 68.68\), \(s(y) = 2.82\)


3.25: Secondary extinction
The secondary extinction is responsible for the loss of intensity occurring when the incident beam crosses a crystal. Each plane of a family ( hkl ) satisfying Laue equations (or Bragg's law ) diffracts the incident beam, and thus subtracts part of the intensity to the incident beam. Successive planes of the same family will then experience a weakening of the incident beam and as a consequence the diffracted beams will result from the positive interference of waves not having the same intensities, as it is instead considered by the kinematical theory.
Secondary extinction is equivalent to an increase of the linear absorption coefficient and is negligible for sufficiently small crystals.


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1.8.2: Reading and Review Questions
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How, why, and to what effect does van der Donck assert the right of the Dutch to possess New Netherland? What assumptions about law and jurisdiction does he make, and why?
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What details of Dutch culture can be inferred by van der Donck’s description of the land, crops, and people in New Netherland? How does van der Donck’s description in this way differ from that of other New World accounts?
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Van der Donck records details of American Indian lives and culture to prevent regret over their inevitable disappearance. According to van der Donck, who will feel this regret? Why?
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Why did the Dutch “name” American Indians the “wilden?” How does van der Donck justify this name as appropriate, even as he acknowledges that different tribes and groups exist, each with their own distinct name?
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Van der Donck recounts an American Indian noting how the Christian God allows himself to be burnt (when a Christian altar and relics catch fire and burn). Van der Donck declares that the American Indian mocked the Christian God. Why does van der Donck recount this event and this so-called mockery? What do you think is van der Donck’s point?


40.13: Blood Flow and Blood Pressure Regulation - Blood Pressure
- Describe the process of blood pressure regulation
Blood Pressure
Blood pressure is the pressure of the fluid (blood) against the walls of the blood vessels. Fluid will move from areas of high to low hydrostatic pressures. In the arteries, the hydrostatic pressure near the heart is very high. Blood flows to the arterioles (smaller arteries) where the rate of flow is slowed by the narrow openings of the arterioles. The systolic pressure is defined as the peak pressure in the arteries during the cardiac cycle; the diastolic pressure is the lowest pressure at the resting phase of the cardiac cycle. During systole, when new blood is entering the arteries, the artery walls stretch to accommodate the increase of pressure of the extra blood. During diastole, the walls return to normal because of their elastic properties.
Blood pressure values are universally stated in millimeters of mercury (mm Hg). The blood pressure of the systole phase and the diastole phase gives the two readings for blood pressure. For example, the typical value for a resting, healthy adult is 120/80, which indicates a reading of 120 mm Hg during the systole and 80 mm Hg during diastole.
Blood Pressure Regulation
Throughout the cardiac cycle, the blood continues to empty into the arterioles at a relatively even rate. However, these measures of blood pressure are not static; they undergo natural variations from one heartbeat to another and throughout the day. The measures of blood pressure also change in response to stress, nutritional factors, drugs, or disease. The body regulates blood pressure by changes in response to the cardiac output and stroke volume.
Cardiac output is the volume of blood pumped by the heart in one minute. It is calculated by multiplying the number of heart contractions that occur per minute (heart rate) times the stroke volume (the volume of blood pumped into the aorta per contraction of the left ventricle). Therefore, cardiac output can be increased by increasing heart rate, as when exercising. However, cardiac output can also be increased by increasing stroke volume, such as if the heart were to contract with greater strength. Stroke volume can also be increased by speeding blood circulation through the body so that more blood enters the heart between contractions. During heavy exertion, the blood vessels relax and increase in diameter, offsetting the increased heart rate and ensuring adequate oxygenated blood gets to the muscles. Stress triggers a decrease in the diameter of the blood vessels, consequently increasing blood pressure. These changes can also be caused by nerve signals or hormones; even standing up or lying down can have a great effect on blood pressure.
Contributions and Attributions
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...ol11448/latest . License : CC BY: Attribution
- Human Physiology/The cardiovascular system. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Human_P...gh_Capillaries . License : CC BY-SA: Attribution-ShareAlike
- artery. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/artery . License : CC BY-SA: Attribution-ShareAlike
- arteriole. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/arteriole . License : CC BY-SA: Attribution-ShareAlike
- vein. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/vein . License : CC BY-SA: Attribution-ShareAlike
- vena cava. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/vena_cava . License : CC BY-SA: Attribution-ShareAlike
- Blood circulation. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Blood_circulation . License : CC BY: Attribution
- OpenStax College, Blood Flow and Blood Pressure Regulation. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...40_04_01ab.png . License : CC BY: Attribution
- Circulatory System en. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ci..._System_en.svg . License : CC BY: Attribution
- OpenStax College, Biology. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...ol11448/latest . License : CC BY: Attribution
- OpenStax College, Biology. October 23, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...ol11448/latest . License : CC BY: Attribution
- Human Physiology/The cardiovascular system. Provided by : Wikibooks. Located at : en.wikibooks.org/wiki/Human_P...gh_Capillaries . License : CC BY-SA: Attribution-ShareAlike
- hydrostatic. Provided by : Wiktionary. Located at : en.wiktionary.org/wiki/hydrostatic . License : CC BY-SA: Attribution-ShareAlike
- cardiac output. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/cardiac%20output . License : CC BY-SA: Attribution-ShareAlike
- stroke volume. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/stroke%20volume . License : CC BY-SA: Attribution-ShareAlike
- Blood circulation. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/Blood_circulation . License : CC BY: Attribution
- OpenStax College, Blood Flow and Blood Pressure Regulation. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...40_04_01ab.png . License : CC BY: Attribution
- Circulatory System en. Provided by : Wikipedia. Located at : en.Wikipedia.org/wiki/File:Ci..._System_en.svg . License : CC BY: Attribution
- OpenStax College, Blood Flow and Blood Pressure Regulation. October 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m44806/latest...e_40_04_03.jpg . License : CC BY: Attribution
Key Points
- Normal blood pressure for a healthy adult is 120 mm Hg during systole (peak pressure in the arteries ) and 80 mm Hg during diastole (the resting phase).
- Blood pressure is regulated in the body by changes to the diameters of blood vessels in response to changes in the cardiac output and stroke volume.
- Factors such as stress, nutrition, drugs, exercise, or disease can invoke changes in the diameters of the blood vessels, altering blood pressure.
Key Terms
- cardiac output : the volume of blood being pumped by the heart, in particular by a left or right ventricle in the time interval of one minute
- hydrostatic : of or relating to fluids, especially to the pressure that they exert or transmit
- stroke volume : the volume of blood pumped from one ventricle of the heart with each beat


9.5: South Coast
The South Coast is a region of California, making up roughly the southernmost third of the Californian coast. It refers for the most part to the Southern California coastal counties of Ventura, Los Angeles, Orange, and San Diego due to the cosmopolitan "SoCal" atmosphere and location of major urban coastal centers. Of these counties only the western two-thirds of San Diego, coastal half of Ventura, most of Los Angeles and all of Orange are included.
Geologists consider the northern Channel Islands to be a westward extension of the Santa Monica’s into the Pacific Ocean. The range was created by repeated episodes of uplifting and submergence by the Raymond Fault, which created complex layers of sedimentary rock, some containing fossils of invertebrates and fish. Volcanic intrusions have been exposed, including the poorly named andesitic. "Sandstone Peak", which is the highest point in the range at 3,111 feet (948 m). Malibu Creek, which eroded its own channel while the mountains were slowly uplifted, bisects the mountain range.
However, some sources include the western part of Riverside, and southwestern part of San Bernardino Counties, and the northwestern corner of Baja California, because of their proximity to the Pacific Coast and because they are in the same bio-region and watershed.
Physical Geography
The Peninsular Ranges are a group of mountain ranges in the Pacific Coast Ranges, which stretch over 900 miles from southern California in the United States to the southern tip of Mexico's Baja California peninsula. They are part of the North American Coast Ranges that run along the Pacific coast from Alaska to Mexico. Elevations range from 500 ft to 11,500 ft (150 m to 3,500 m) and vegetation in these ranges varies from coastal sage scrub to chaparral, and from oak woodland to conifer forest.
The Peninsular Ranges of southern California include the Santa Ana Mountains, Santa Monica Mountains, San Jacinto Mountains, and the Laguna Mountains. The Peninsular Ranges of Baja California include the Sierra Juarez, Sierra San Pedro Martir, Sierra de la Giganta, and Sierra de la Laguna. These ranges run from north to south.
The Santa Ana Mountains are the largest natural landscape along the coast of southern California. These mountains peak at about 5,689 feet, on Santiago Peak. This range starts in the north, in the Corona area heading southeast of the Puente Hills region.
The Santa Monica Mountains are approximately 40 miles (64 km) east-west from the Hollywood Hills in Los Angeles to Point Mugu in Ventura County. The western mountains, separating the Conejo Valley from Malibu, suddenly end at Mugu Peak as the rugged, nearly impassible shoreline gives way to tidal lagoons and coastal sand dunes of the alluvial Oxnard Plain. The mountain range contributed to the isolation of this vast coastal plain before regular transportation routes reached western Ventura County. The eastern mountains form a barrier between the San Fernando Valley and the Los Angeles Basin, separating "the Valley" on the north and west-central Los Angeles on the south. The Santa Monica Mountains are parallel to the Santa Susana Mountains, which are located directly north of the mountains across the San Fernando Valley.
The San Jacinto Mountains are in the desert areas in the north and east side of southern California. They peak at about 10,833 feet. They run from the San Bernardino Mountains southeast to the Santa Rosa Mountains. This mountain range is the northernmost part of the Peninsular Range.
The Santa Rosa Mountains are located at the southern end of the San Jacinto Mountains, where they connect to it. The range extends for approximately 30 miles (48 km) through Riverside, San Diego and Imperial counties, along the western side of the Coachella Valley, where they bound the Anza-Borrego portion of the Colorado Desert. The highest peak in the range is Toro Peak (8,717 feet).
The Laguna Mountains are in the eastern part of San Diego County. They range northwest to southeast for approximately 20 miles and peak at Cuyapaipe Mountain (6,378 feet). These mountains extend northwest about 35 mi (56 km) from the Mexican border at the Sierra de Juárez. The Sonora Desert lies to the east and the Santa Rosa Mountains are to the northwest.
Case Study – Balboa Island
Millions of years ago, powerful geological forces shaped the landscape we see today. The relentless pounding of the Pacific Ocean and the deposition of sediment by the Santa Ana River gradually formed a submerged sandbar off the coast of Newport Beach. This underwater feature, continuously evolving under the influence of tides and currents, eventually emerged above the waterline sometime between 1825 and 1865 as a small mudflat, that was surrounded by swampland. This newly formed sandbar laid the foundation for what would become Balboa Island.
Around 1860, James McFadden, a prominent Newport Beach landowner, envisioned transforming the undeveloped sandbar into a vibrant resort destination. In the early 1900s, W.S. Collins, a real estate developer, saw the potential of McFadden's vision. He purchased a significant portion of the land and began dredging the harbor to create a channel and further define the island's shape. This period also saw the construction of the Pacific Electric Red Car line, connecting Balboa Island to Los Angeles and making it a more accessible getaway for city dwellers.
The early 20 th century marked a golden age for Balboa Island. The iconic Balboa Pavilion, originally built as a Victorian bathhouse, quickly transformed into a renowned entertainment venue, hosting dances, concerts, and other social gatherings. The construction of the Balboa Fun Zone, with its Ferris wheel, arcade games, and carnival atmosphere, further solidified the island's reputation as a premier leisure destination. The 1920s saw the development of unique "Nutcracker" houses, small, colorful cottages designed to maximize space. These charming dwellings remain a hallmark of the island's character.
During World War II, the island played a vital role in the war effort. The Balboa Yacht Basin served as a training ground for the US Navy, and the island's residents proudly supported the war cause. Following the war, Balboa Island experienced continued growth and development, maintaining its status as a beloved seaside escape.


6.3: Simple Interest
- Compute simple interest.
- Understand and compute future value.
- Compute simple interest loans with partial payments.
- Understand and compute present value.
There is truth in the phrase “You need to have money to make money.” In essence, if you have money to lend, you can lend it at a cost to a borrower and make money on that transaction.
When money is borrowed, the person borrowing the money (borrower) typically has to pay the person or entity that lent the money (the lender) more than the amount of money that was borrowed. This extra money is the interest that is to be paid. Interest is sometimes referred to as the cost to borrow, the cost of the loan, or the finance cost.
This idea also applies when someone deposits money in a bank account or some other form of investment. That person is essentially lending the money to the bank or company. The money earned by the depositor is also called interest. The interest is typically based on the amount borrowed, or the principal .
The pairing of borrower and lender can take various forms. The borrower may be a consumer using a credit card or taking out a loan from a bank, the lender. Companies also borrow from lending banks. Someone who invests in a company’s stock is the lender in this case; the company is essentially the borrower.
In this section, we examine the basic building block of interest paid on loans and borrowed credit and also the returns on investments like bank accounts, simple interest.
Compute Simple Interest
Let’s get some terminology understood. Interest to be paid by a borrower is often expressed as an annual percentage rate , which is the percent of the principal that is paid as interest for each year the money is borrowed. This means that the more that is borrowed, the more that must be paid back. Sometimes, the interest to be paid back is simple interest , which means that the interest is calculated on the amount borrowed only.
The length of time until the loan must be paid off is the term of the loan. The date when the loan must be paid off is when the loan is due . The day that the loan is issued is the origination date . We’ll put this terminology to use in the following examples. Note that in this section we will use letters, called variables, to represent the different parts of the formulas we’ll be using. This will help keep our formulas and calculations manageable.
Simple Interest Loans with Integer Year Terms
Calculating simple interest is similar to the percent calculations we made in Understanding Percent and Discounts, Markups, and Sales Tax, but must be multiplied by the term of the loan (in years, if dealing with an annual percentage rate).
The simple interest,
, to be paid on a loan with annual interest rate
for a number of years (term of the loan)
, with principal
, is found using
When the annual interest rate, the principal, and the number of years that the money is borrowed is known, the interest to be paid can be found and from there the total to be repaid can be calculated.
Be aware, interest paid to a lender is almost uniformly rounded up to the next cent.
Calculate the simple interest to be paid on a loan with the given principal, annual percentage rate, and number of years. Then, calculate the loan payoff amount.
- Principal = $4,000, annual interest rate = 5.5%, and number of years = 4
- Principal = $14,800, annual interest rate = 7.9%, and number of years = 7
- Answer
-
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Substitute the principal
= $4,000, the decimal form of the annual interest rate
= 0.055, and number of years
= 4 into the formula for simple interest, and calculate.
880 I .= P × r × t = 4,000 × 0.055 × 4 = 880
The simple interest, or cost of the loan, to be paid on the loan is $880.
The loan payoff amount, or the total to be repaid, is880 T , or $4,880.00.= P + I = 4,000 + 880 = 4,880 -
Substitute the principal
= $14,800, the decimal form of the annual interest rate
= 0.079, and number of years
= 7 into the formula for simple interest, and calculate.
.
The simple interest, or cost to borrow, to be paid on the loan is $8,184.40.
The loan payoff amount, or the total to be repaid, is , or $22,984.40.
-
Substitute the principal
= $4,000, the decimal form of the annual interest rate
= 0.055, and number of years
= 4 into the formula for simple interest, and calculate.
Calculate the simple interest to be paid on a loan with the given principal, annual percentage rate, and number of years. Then calculate the loan payoff, or total to be repaid.
Principal \(P\) = $6,700, annual interest rate \(r\) = 11.99%, and number of years \(t\) = 3
Principal \(P\) = $25,800, annual interest rate \(r\) = 6.9%, and number of years \(t\) = 5
Riley runs an auto repair shop, and needs to purchase a new brake lathe, which costs $11,995. She takes out a two-year, simple interest loan at an annual interest rate of 14.9%. How much interest will she pay and how much total will she repay on the loan?
- Answer
-
Step 1. Determine the variables, or parts of the formula. The principal is the cost of the brake lathe, so = $11,995. The interest rate Riley pays is 14.9%, or = 0.149 in decimal form. The length of the loan is two years, so = 2. We are first asked to find , the interest Riley will pay.
Step 2. Substitute the known variables into the formula for simple interest
P × r × t I and solve for .= P × r × t From Step 1 we have .
This tells us that the simple interest, or cost to borrow, to be paid on the loan is $3,574.51.
Step 3. Use the formula to determine the total amount Riley will repay, .
The total to be repaid is , or $15,569.51.
Beth is the owner of a small retail store in downtown St. Louis. The windows in the storefront need replacing, so she needs to take out a $9,500 loan to get the repairs done. The rate she secures is 9.25% and the term of the loan is one year. How much interest will she pay and how much total will she repay on the loan?
Simple Interest Loans with Other Lengths of Terms
In the previous example and Your Turn exercise, the loans were paid back in one payment after an integer number of years. However, there are also loans lasting a length of time not equal to an integer number of years (like 1, 2, or 3 years or more), but in a number of months (like 4 months, 18 months, and so on). What model would apply to these situations?
When the loan is paid back after a term that is not an integer number of years but is instead a number of months, the term of the loan, or time, , is expressed as a fraction of the year. So for a 2-month loan, the time, in years, is 2/12 = 1/6. For a 5-month loan, the time in years is 5/12. For an 18-month term, the term in years is 18/12 = 1.5.
Abeje needs a loan to purchase equipment for the gym she is going to open. She visits the bank and secures a 4-month loan of $20,000. Her annual percentage rate is 6.75%. How much interest will Abeje pay and what is her loan payoff amount?
- Answer
-
Abeje’s loan is for $20,000, so her principal is = 20,000. The interest rate Abeje will pay is 6.75%, or = 0.0675 in decimal form. The length of the loan is 4 months, so
4 12 t . Substituting these in the formula for simple interest, we find her interest to be= 4 12 450 I = P × r × t = 20,000 × 0.0675 × 4 / 12 = 450 The simple interest, or cost to borrow, to be paid on the loan is $450.00.
The payoff is
450 T , or $20,450.00.= P + I = 20,000 + 450 = 20,450
Samuel needs to borrow $8,400 to pay for repairs to his small manufacturing facility. He manages to get a simple interest loan at 17.33%, to be paid after 6 months. How much interest will Samuel pay and what is Samuel’s loan payoff amount?
Those examples dealt in months. However, some loans are for days only (45 days, 60 days, 120 days). In such cases, we find the daily interest rate. The fraction we will use for the daily interest rate is the interest rate (as a decimal) divided by 365. This may be referred to as Actual/365. In order to find the term of the loan, divide the number of days in the term of the loan by 365.
To determine the interest,
, on a loan with term
expressed in days, with principal of
, and interest rate in decimal form of
, calculate
Alternately, the above formula is equivalent to
It seems reasonable to use 365 as the number of days in the year, since there are 365 days in most years. However, sometimes, banks have used (and continue to use) 360 as the number of days in a year. They may also treat all months as if they have 30 days. These differences lead to (sometimes small) differences in how much interest is paid. Since the number of days is in the denominator, a smaller denominator (360) will result in larger numbers (interest) that is 365 is used for the denominator. See this page from ACRE for a comparison.
David plans to move his family from Raleigh, North Carolina to Tempe, Arizona. His company will reimburse (pay after the move) David for the move. David does research and determines that movers will cost $5,600 to move his family’s belongings to Tempe. He takes out a simple interest, 45-day loan at 11.75% interest to pay this cost. How much interest will be paid on this 45-day loan, and what is David’s loan payoff amount?
- Answer
-
This loan is in terms of days, so we will use the formula
P × r 365 × t I , where t is the number of days and is the annual interest rate.= P × r 365 × t The principal for the loan is the moving cost, or = 5,600. The annual interest rate that David will pay is 11.75%, which in decimal is 0.1175. The length of time for the loan is 45 days, so = 45.
Substituting these values into the formula and calculating, we find that the interest to be paid is , or $81.13 (remember, interest is almost always rounded up to the next cent).
The payoff for the loan is $5,681.13.
Heather runs a silk screen t-shirt shop, and seeks a short-term loan to pay for new inventory (paints, blades, shirts). They secure a $3,700 simple interest loan for 60 days, at an annual rate of 18.99%. How much will borrowing the money cost Heather, and what is her loan payoff amount?
Understand and Compute Future Value
Money can be invested for a specific amount of time and earn simple interest while invested. The terminology and calculations are the same as we’ve already seen. However, instead of the total to be paid back, the investor is interested in the total value of the investment after the interest is added. This is called the future value of the investment.
The future value,
In the following, determine how much interest was earned on the investment and the future value of the investment, if the investment yields simple interest.
- Principal is $1,000, annual interest rate is 2.01%, and time is 5 years
- Principal is $5,000, annual interest rate is 1.85%, and time is 30 years
- Principal is $10,000, annual interest rate is 1.25%, and time is 18 months
- Principal is $7,000, annual interest rate is 3.26%, and time is 100 days
- Answer
-
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The principal is
= $1,000, the annual interest rate, in decimal form, is 0.0201, and the term is 5 years, or
= 5. Since the term is an integer number of years, the interest earned on the investment is
, or the interest earned was $100.50.
To find the future value, we use the formulaI F V . Substituting the values and calculating, we find the future value of the investment to be . The future value of the investment at the end of 5 years is $1,100.50.= P + I
Notice that the future value could have been calculated directly withP × r × t F V = P + P × r × t -
The principal is
= $5,000, the annual interest rate, in decimal form, is 0.0185, and the term is 30 years, or
= 30. Since the term is an integer number of years, the interest earned on the investment is
775 I , or the interest earned was $2,775.00. To find the future value, we use the formula= P × r × t = 5,000 × 0.0185 × 30 = 2,775 I F V . Substituting the values and calculating, we find the future value of the investment to be= P + I 775 F V . The future value of the investment at the end of 30 years is $7,775.00.= P + I = 5,000 + 2,775 = 7,775 -
The principal is
= $10,000, the annual interest rate, in decimal form, is 0.0125, and the term is 18 months. Since the term is in months, we have to write the months in terms of years. For 18 months, we use 18/12 as
. The interest earned on the investment is
, or the interest earned was $187.50. To find the future value, we use the formula
I F V . Substituting the values and calculating, we find the future value of the investment to be . The future value of the investment at the end of 18 months is $10,187.50.= P + I -
Principal is $7,000, annual interest rate is 3.26%, and time is 100 days. The principal is
= $7,000, the annual interest rate, in decimal form, is 0.0326, and the term is 100 days. Since the term is in days, we have to write the time using actual/365, or
= 100/365. The interest earned on the investment is
, or the interest earned was $62.52. To find the future value, we use the formula
I F V . Substituting the values and calculating, we find the future value of the investment to be . The future value of the investment at the end of 100 days is $7,062.52.= P + I
-
The principal is
= $1,000, the annual interest rate, in decimal form, is 0.0201, and the term is 5 years, or
= 5. Since the term is an integer number of years, the interest earned on the investment is
, or the interest earned was $100.50.
In the following, determine how much interest was earned on the investment and the future value of the investment if the investment yields simple interest.
Principal is $4,500, annual interest rate is 1.88%, and time is 3 years
Principal is $2,000, annual interest rate is 2.03%, and time is 10 years
Principal is $120,000, annual interest rate is 3.1%, and time is 100 days
Principal is $4,680, annual interest rate is 1.55%, and time is 42 months
You may have noticed that for these problems, the future value was rounded down. When the future value is paid, the amount is typically rounded down.
A certificate of deposit (CD) is a savings account that holds a single deposit (the principal) for a fixed term at a fixed interest rate. Once the term of the CD is over, the CD may be redeemed (cashed in or withdrawn) and the owner of the CD receives the original principal plus the interest earned. The deposit often cannot be withdrawn until the term is up; if it can be withdrawn early, there is often a penalty imposed to do so.
Jonas deposits $2,500 in a CD bearing 3.25% simple interest for a term of 3 years. When he redeems his CD at the end of the 3 years, how much will he receive?
- Answer
-
This is a future value example. We know that = $2,500 is the amount deposited. The annual simple interest rate in decimal form is = 0.0325. The term of the investment is = 3 years.
Substituting those values into the future value formula, we have .
When the CD is redeemed, Jonas will receive $2,743.75.
Mia deposits $4,900 in a CD bearing 3.95%. The CD term is 7 years. When she redeems the CD, how much will Mia receive?
The reason CD (certificate of deposit) rates look so small is because they are extremely safe investments. Though overall interest rates for CDs change over time and individual returns vary with the terms of the CD, investors are offered predictable interest income for their investments.
To investigate this yourself, search online to determine the strengths and weaknesses of CDs (investopedia.com offers good, basic information on investing). Then, online, identify five national banks and two local banks who offer CDs.
- Track the interest rates for the CDs at various terms (1 year, 3 years, 5 years) for each of the banks you found that offer CDs.
- Calculate the amount of interest earned for a $10,000 deposit for each CD at each of the terms.
- Compare the results from the various banks, CDs, and terms and decide which is the best investment. You may want to consider both the length of time that the money is locked up, and the return.
Paying Simple Interest Loans with Partial Payments
In every example above, there was one payment for the loan, or one withdrawal for the investment. However, for many loans (house, car, in-ground swimming pool), the loan will be paid back in two or more payments. Such a payment is called a partial payment , because they only pay off part of the loan.
When a partial payment is made, some of the payment pays for the principal, but the rest of the payment pays for interest on the principal. When making the first partial payment, the interest is calculated on the principal for the time between the origination date of the loan and the date of the payment. If another partial payment is made, the interest is calculated based on the remaining principal and the time between the previous partial payment and the current partial payment date.
- A simple interest loan for $6,500 is taken out at 12.6% annual percentage rate. A partial payment is made 45 days into the loan period. How much of the partial payment will be for interest?
- A simple interest loan for $13,700 is taken out at 6.55% annual interest rate. A partial payment is to be made after 60 days. How much of the partial payment will be for interest?
- Answer
-
-
To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 45 days.
The principal is $6,500. The annual interest rate, in decimal form, is 0.126.
The interest paid for 45 days is found by substituting the values for principal , rate , and time into the formulaP × r 365 × t I .= P × r 365 × t
Calculating, we haveP × r 365 × t = 6,500 × 0.126 365 × 45 = 100.9726 . Rounding up, the portion of the partial payment that will be paid for interest is $100.98.I = P × r 365 × t = 6,500 × 0.126 365 × 45 = 100.9726 -
To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 60 days.
The principal is $13,700. The annual interest rate, in decimal form, is 0.0655.
The interest paid for those 60 days is found by substituting those values into the formulaP × r 365 × t I . Calculating, we have= P × r 365 × t P × r 365 × t = 13,700 × 0.0655 365 × 60 = 147.5096 . Rounding up, the portion of the partial payment that will be paid for interest is $147.51.I = P × r 365 × t = 13,700 × 0.0655 365 × 60 = 147.5096
-
To find the interest paid in this partial payment, we calculate the interest on the principal for the time between the origination of the loan and the payment day, or 45 days.
A simple interest loan for $50,000 is taken out at 5.15% annual percentage rate. A partial payment is made 120 days into the loan period. How much of the partial payment will be for interest?
A simple interest loan for $8,500 is taken out at 9.9% annual interest rate. A partial payment is to be made after 75 days. How much of the partial payment will be for interest?
Remaining Balance
The previous examples demonstrated how to determine the interest paid in a partial payment. Using this, we can determine the remaining balance after a partial payment.
Step 1: determine the amount of the payment, , that is applied to interest, .
Step 2: subtract the amount paid in interest from the payment, . This is the amount applied to the balance.
Step 3: subtract the amount applied to the balance (the value obtained in Step 2) from the balance of the loan, . This is the remaining balance after the partial payment.
- A simple interest loan for $45,500 is taken out at 11.8% annual percentage rate. A partial payment of $20,000 is made 50 days into the loan period. After this payment, what will the remaining balance of the loan be?
- A simple interest loan for $150,000 is taken out at 5.85% annual percentage rate. A partial payment of $50,000 is made 70 days into the loan period. After this payment, what will the remaining balance of the loan be?
- Answer
-
-
The principal is $45,500, which will be treated as the balance,
, of the loan. The annual simple interest rate, in decimal form, is 0.118. The time is
= 50 days.
Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formulaP × r 365 × t I and calculate. Calculating, the amount of the payment that is applied to interest is . Rounding up, we have = $735.48.= P × r 365 × t
Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $2,000. The amount that is applied to the balance is .
Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or .
The remining balance after the partial payment is $26,235.48. -
The principal is $150,000, which will be treated as the balance,
, of the loan. The annual simple interest rate, in decimal form, is 0.0585. The time is
= 70 days.
Step 1: Determine the amount of the partial payment that is applied to interest. To find this, substitute the values above into the formulaP × r 365 × t I and calculate. Calculating, the amount of the payment that is applied to interest is . Rounding up, we have = $1,682.88.= P × r 365 × t
Step 2: The amount of the payment that is to be applied to the balance of the loan is partial payment minus the amount of the partial payment that is applied to the interest. The payment is $50,000. The amount that is applied to the balance is .
Step 3: The remaining balance is found by subtracting the amount applied to the balance from the previous balance, or .
The remining balance after the partial payment is $101,682.88.
-
The principal is $45,500, which will be treated as the balance,
, of the loan. The annual simple interest rate, in decimal form, is 0.118. The time is
= 50 days.
A simple interest loan for $1,400 is taken out at 12.5% annual percentage rate. A partial payment of $700 is made 20 days into the loan period. After this payment, what will the remaining balance of the loan be?
A simple interest loan for $23,000 is taken out at 7.25% annual percentage rate. A partial payment of $10,000 is made 40 days into the loan period. After this payment, what will the remaining balance of the loan be?
Loan Payoff
Finally, we will determine the amount to be paid at the end of the loan. To do so, we apply the formula for the loan payoff to the remaining balance. However, the length of time for that remaining balance is the time between the partial payment and the day the loan is paid off.
Step 1: Determine the remaining balance after the partial payment.
Step 2: Calculate the number of days between the partial payment and the date the loan is paid off. This will be the time in the payment formula.
Step 3:
Calculate the amount to be paid at the end of the loan, or the payoff amount, using
Laura takes out an $18,400 loan for 120 days at 17.9% simple interest. She makes a partial payment of $7,500 after 45 days. What is her payoff amount at the end of the loan?
- Answer
-
The initial balance, or principal, of her loan is $18,400. The interest rate in decimal form is 0.179. Her partial payment of $7,500 is made after 45 days. Using these values, we can determine how much of the partial payment is applied to the balance. From there, we can determine her final loan payoff after 120 days.
Step 1: Determine the remaining balance after the partial payment. Using the partial payment process outlined in the previous example, we first find that the amount of the partial payment that is applied to the balance. Their interest paid in the partial payment is
P × r 365 × t = 18,400 × 0.179 365 × 45 = 406.0603 , or $406.07 (remember to round up!). Using this and that the loan amount was for $18,400, the remaining balance on the loan after the partial payment is .I = P × r 365 × t = 18,400 × 0.179 365 × 45 = 406.0603 Step 2: The number of days between the partial payment and the date that the loan is to be paid off is 120 – 45 = 75. This means that the time between the partial payment and the final payment is 75 days.
Step 3: To calculate the payoff amount, use
P × r 365 × t payoff , with = $11,306.07 (the remaining balance), = 75 (from Step 2) and = 0.179. The payoff amount, then, is . Rounding up, the payoff amount is $11,721.92.= P + P × r 365 × t
Paola takes out a 75-day loan for $3,500.00. Her interest rate is 11.2%. If she makes a partial payment of $1,250.00 after 30 days, what will her payoff be at the end of the loan?
Repeated Partial Payments
Car loans and mortgages (loans for homes) are paid off through repeated partial payments, most often monthly payments. Since car loans are often 3 to 6 years, and mortgages 15 to 30 years, calculating each individual monthly payment one at a time is time consuming and tedious. Even a 3-year loan would involve applying the above steps 36 times! Fortunately, there is a formula for determining the amount of each partial payment for monthly payments on a simple interest loan.
The amount of monthly payments, , for a loan with principal , monthly simple interest rate (in decimal form), for number of months is found using the formula . The monthly interest rate is the annual rate divided by 12. The number of months is the number of years times 12.
Desiree buys a new car, by taking a loan out from her credit union. The balance of her loan is $27,845.00. The annual interest rate that Desiree will pay is 7.3%. She plans to pay this off over 4 years. How much will Desiree’s monthly payment be?
- Answer
-
To use the formula for monthly payments, we need the principal, the interest rate, and the number of years. The principal is $27,845. The annual rate, in decimal form, is 0.073. Dividing 0.073 by 12 gives the monthly interest rate . She takes the loan out for 4 years, which is
48 t months. Substituting these values into the formula, , we calculate:= 12 × 4 = 48 Using the formula and rounding up to the next cent, we see that Desiree’s monthly payment will be $670.67.
Russell buys a new car, by taking a loan out from the dealership. After all of the discussions are over, he finances (gets a loan for) $23,660. The annual interest rate that Russell will pay is 4.76%. He plans to pay this off over 5 years. How much will Russell’s monthly payment be?
The calculation of payments is long, and involves many steps. However, most spreadsheet programs, including Google Sheets, have a payment function. In Google Sheets, that function is PMT. To find the payment for an installment loan (like for a car), you need to enter the interest rate per period, the number of payments, and the loan amount. From Your Turn 6.36, the rate was 0.0476/12, the number of payments was 60, and the loan amount was $23,660. In Google Sheets, select any cell and enter the following:
=PMT(0.0476/12,60,23660)
And click the enter key. Immediately, in the cell you selected, the payment of $443.90 appears, though with a negative sign. The negative sign indicates it is a payment out of an account. Since we want to know the payment amount, we ignore the negative sign. The result, with the formula in the formula bar, is shown in Figure 6.5.
In general, to use the PMT function in Google Sheets, enter
=PMT( /12, *12, )
where is the annual interest rate, is the number of years, and is the principal of the loan.
Understand and Compute Present Value for Simple Interest Investments
When finding the future value of an investment, we know how much is deposited, but we have no idea how much that money will be worth in the future. If we set a goal for the future, it would be useful to know how much to deposit now so an account reaches the goal. The amount that needs to be deposited now to hit a goal in the future is called the present value .
The present value,
Note: Present value, in this calculation, is always rounded up. Otherwise, future value may fall short of the target future value.
Understanding what this tells you is important. When you find the present value, that is how much you need to invest now to reach the goal
Compute the present value of the investment described. Interpret the result.
-
F V = $10,000, = 15 years, annual simple interest rate of 5.5%F V -
F V = $150,000, = 20 years, annual simple interest rate of 6.25%F V -
F V = $250,000, = 486 months, annual simple interest rate of 4.75%F V
- Answer
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-
The future value is
F V = $10,000. The time of the investment is in years, so = 15. The annual, simple interest rate is 5.5%, which in decimal form is 0.055. We substitute those values into the formula and calculate. . Rounding up, we see that the present value of $10,000 invested at a simple annual interest rate of 5.5% for 15 years is $5,479.46. This means that $5,479.46 needs to be invested so that, after 15 years at 5.5% interest, the investment will be worth $10,000.F V -
The future value is
F V = $150,000. The time of the investment is in years, so = 20. The annual, simple interest rate is 6.25%, which in decimal form is 0.0625. We substitute those values into the formula and calculate. . Rounding up, we see that the present value of $150,000 invested at a simple annual interest rate of 6.25% for 20 years is $66,666.67. This means that $66,666.67 needs to be invested so that, after 20 years at 6.25% interest, the investment will be worth $150,000.F V -
The future value is
FV = $250,000. The time of the investment is 486 months. This needs to be converted to years. To do so, divide the number of months by 12, giving , so = 40.5 years. The annual, simple interest rate is 4.75%, which in decimal form is 0.0475. We substitute those values into the formula and calculate. . Rounding up, we see that the present value of $250,000 invested at a simple annual interest rate of 4.75% for 486 months is $129,954.52. This means that $129,954.52 needs to be invested so that, after 486 months at 4.75% interest, the investment will be worth $150,000.FV
-
The future value is
Compute the present value of the investment described. Interpret the result.
\(FV\) = $25,000, \(t\) = 10 years, annual simple interest rate of 7.5%
\(FV\) = $320,000, \(t\) = 35 years, annual simple interest rate of 6.5%
\(FV\) = $90,000, \(t\) = 270 months, annual simple interest rate of 3.75%
Beatriz will invest some money in a CD that yields 3.99% simple interest when invested for 30 years. How much must Beatriz invest so that after those 30 years, her CD is worth $300,000?
- Answer
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Beatriz needs to know how much to deposit now so that her CD is worth $300,000 after 30 years. This means she needs to know the present value of that $300,000. The time is 30 years and the annual simple interest rate, in decimal form, is 0.0399. Using that information and the formula for present value, we calculate the present value of that $300,000. . Rounding up, Beatriz needs to invest $136,549.85 so that she has $300,000 in 30 years.
Kentaro will invest some money in a CD that yields 2.75% simple interest when invested for 5 years. How much must Kentaro invest so that after those 5 years his CD is worth $12,000?
Check Your Understanding
- What is interest?
- What is the principal in a loan?
- Calculate the simple interest to be paid for a 6-year loan with principal $1,500.00 and annual interest rate of 12.99%
- A simple interest loan for $24,200 is taken out at 10.55% annual percentage rate. A partial payment of $13,000 is made 25 days into the loan period. After this payment, what will the remaining balance of the loan be?
- Find the monthly payment for a $9,800.00 loan at a 13.8% interest for 4 years.
- Find the present value of an investment with future value $30,000 with a simple interest rate of 3.75% invested for 10 years.


6.3: Reading- Characteristics of the Crust
How does a loaf of bread resemble Earth?
A loaf of homemade bread could almost resemble Earth. The raised parts of the crust are the continents and the depressed parts are the oceans. The inside is gooier than the brittle exterior, but it’s still solid. How is a loaf of bread not like Earth?
Earth’s outer surface is its crust, a cold, thin, brittle outer shell made of rock. The crust is very thin relative to the radius of the planet. There are two very different types of crust, each with its own distinctive physical and chemical properties, which are summarized in Table 1.
| Crust | Thickness | Density | Composition | Rock Types |
|---|---|---|---|---|
| Oceanic | 5–12 km (3–8 mi) | 3.0 g/cm 3 | Mafic | Basalt and gabbro |
| Continental | Avg. 35 km (22 mi) | 2.7 g/cm 3 | Felsic | All types |
Oceanic Crust
Oceanic crust is composed of mafic magma that erupts on the seafloor to create basalt lava flows or cools deeper down to create the intrusive igneous rock gabbro (Figure 1).
Sediments, primarily mud and the shells of tiny sea creatures, coat the seafloor. Sediment is thickest near the shore, where it comes off the continents in rivers and on wind currents.
The oceanic crust is relatively thin and lies above the mantle. The cross section of oceanic crust in Figure 2 shows the layers that grade from sediments at the top to extrusive basalt lava, to the sheeted dikes that feed lava to the surface, to deeper intrusive gabbro, and finally to the mantle.
Continental Crust
Continental crust is made up of many different types of igneous, metamorphic, and sedimentary rocks. The average composition is granite, which is much less dense than the mafic rocks of the oceanic crust (Figure 3). Because it is thick and has relatively low density, continental crust rises higher on the mantle than oceanic crust, which sinks into the mantle to form basins. When filled with water, these basins form the planet’s oceans.
Summary
- Oceanic crust is thinner and denser than continental crust.
- Oceanic crust is more mafic, continental crust is more felsic.
- Crust is very thin relative to Earth’s radius.
Interactive Practice
Visit Annenberg Learner’s Dynamic Earth and click on the crust to answer the questions below:
- Describe the crust.
- Where is the crust the thickest?
Contributors and Attributions
- Earth's Crust. Provided by : CK-12. Located at : http://www.ck12.org/earth-science/Earths-Crust/lesson/Earths-Crust-HS-ES/?referrer=featured_content . License : CC BY-NC: Attribution-NonCommercial