 3 x 90 6 + 0 x 200 P(0 X 200)= 0 (b) 200 P(0 X 200)= 0 (c) P(X>800)= 800 (d) P(X>800)= 800 1 e 1000 t/1000 1 e 1000 t/1000 1 e 1000 t/1000 1 e 1000 t/1000 dt= e dt= e dt=lim e x dt=lim x e 6 16 =5.3 3 1 1 t/5 1 1 7. We need to find m so that f (t)dt= lim e dt= lim ( 5)e 2 x 2 x 5 m m 5 1 1 m/5 1 m/5 1 ( 1)(0 e )= e = m/5=ln m= 5ln =5ln 2 3.47 min. 2 2 2 2 8. (a) (a) 10 1 3 1 2 x+ x 60 4 t/1000 200 0 = e t/1000 200 0 t/1000 x = e = m 1 2 +1 0.181 1/5 =0+e =0+e 800 x 1/5 +1 0.181 4/5 800 t/1000 x t/5 4/5 0.449 0.449 3 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability x 1 1 1 t/1000 (b) We need to find m so that f (t)dt= lim e dt= 2 x 2 m m 1000 1 1 m/1000 1 0+e = m/1000=ln m= 1000ln =1000ln 2 693.1 h. 2 2 2 lim e t/1000 x = m x 1 2 9. We use an exponential density function with =2.5 min. x (a) P(X>4)= f (t)dt=lim x 4 4 1 e 2.5 2 (b) P(0 X 2)= f (t)dt= e t/2.5 e t/2.5 x =0+e = e 4/2.5 4 x t/2.5 2 0 dt=lim 0.202 2/2.5 +1 0.551 0 (c) We need to find a value a so that P(X a)=0.02, or, equivalently, P(0 X a)=0.98 a f (t)dt=0.98 e t/2.5 a =0.98 0 0 e a/2.5 +1=0.98 e a/2.5 =0.02 a/2.5=ln 0.02 1 =2.5ln 50 9.78 min 10 min. The ad should say that if you aren’t served within 10 50 minutes, you get a free hamburger. a= 2.5ln 73 1 10. (a) With =69 and =2.8, we have P(65 X 73)= 2.8 2 exp ( x 69 ) 2 2 dx 0.847 2 2.8 65 (using a calculator or computer to estimate the integral). (b) P(X>6 feet )=P(X>72 inches )=1 P(0 X 72) 1 0.858=0.142, so 14.2% of the adult male population is more than 6 feet tall. 1 11. P(X 10)= 10 4.2 2 exp ( x 9.4 ) 2 2 dx. To avoid the improper integral we approximate it 2 4.2 100 1 by the integral from 10 to 100. Thus, P(X 10) 4.2 2 exp ( x 9.4 ) 2 2 dx 0.443 (using a 2 4.2 calculator or computer to estimate the integral), so about 44 percent of the households throw out at least 10 lb of paper a week. Note : We can’t evaluate 1 P(0 X 10) for this problem since a significant amount of area lies to the left of X=0. 10 12. (a) 4 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 480 1 12 2 P(0 X 480)= ( x 500 ) exp 2 2 dx 0.0478 (using a calculator or computer to 2 12 estimate the integral), so there is about a 4.78% chance that a particular box contains less than 480 g of cereal. (b) We need to find so that P(0 X<500)=0.05. Using our calculator or computer to find P(0 X 500) for various values of, we find that if =519.73, P=0.05007 ; and if =519.74, P=0.04998. So a good target weight is at least 519.74 g. 0 +2 13. P( 2 X.. 1 2 +2 )= 2 exp 2 gives us 14. Let f (x)= bx 2 bx 2 = 1 2 2 0 ce e if x<0 if x 0 cx t 2 / 2( 2 dx. Substituting t= 2 2 1 dt)= 2 e t 2 / 2 dt x and dt= 1 dx 0.9545 2 where c=1/. By using parts, tables, or a CAS, we find that / b ) ( bx 1) dx= ( e / b ) ( b x 2bx+2 ) (1): xe dx= (2): x e Now. { 2 (x ) (x ( e bx 2 bx 3 2 2 0 2 ) f (x)dx= (x 2 ) f (x)dx+ (x 2 ) f (x)dx 0 t = 0+lim c (x t 2 )e cx t dx=c lim t 0 ( x2e cx 2x cx e + 2 e cx ) dx 0 Next we use (2) and (1) with b= ctoget 2 =clim t e c cx 3 ( c x +2cx+2) 2 2 2 e c cx 2 ( cx 1)+ 2 e cx c t 0 Using l’Hospital’s Rule several times, along with the fact that =1/c, we get 5 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability 2 =c 2 0 c 15. (a) First p(r)= 4 a 3 2 3 + 2 c 1 c + 2 1 c 2 1 c 1 =c c 1 = 3 c = 2 1 = c 2r/a 0 r e 0 for r 0. Next, 0 4 p(r)dr= a 3 2 2r/a 0 r e dr= 0 0 2 bx 4 t 2r/a 2 0 r e lim 3 a t 0 dr 0 bx 3 2 2 By using parts, tables, or a CAS, we find that x e dx=(e /b )(b x 2bx+2). ( * ) a 4 Next, we use ( * ) (with b= 2/a ) and l’Hospital’s Rule to get 0 a 3 0 8 3 ( 2) =1. This satisfies the 0 second condition for a function to be a probability density function. (b) Using l’Hospital’s Rule, 4 r lim 3 a r 0 2 2r/a e = 0 4 2r lim 3 a r 0 2r/a 0 ( 2/a ) e 0 = 2 2 lim 2 a r 0 2r/a ( 2/a ) e =0. 0 0 To find the maximum of p, we differentiate: / p (r)= 4 a 3 2r/a 2 2 a 0 r e / r=0 or 1= r a 0 (2r) = 4 a 0 0 p (r)=0 2r/a +e 3 2r/a e 0 (2r) 0 r +1 a 0 / r=a. p (r) changes from positive to negative at r=a, so p(r) has its 0 0 0 maximum value at r=a. 0 (c) It is fairly difficult to find a viewing rectangle, but knowing the maximum value from part (b) helps. ( ) p a = 0 4 a 3 0 2 ae 0 2a /a 0 0 = 4 e a 2 9,684,098,979 0 6 Stewart Calculus ET 5e 0534393217;8. Further Applications of Integration; 8.5 Probability With a maximum of nearly 10 billion and a total area under the curve of 1, we know that the ‘‘hump’’ in the graph must be extremely narrow. 4a r 0 4 (d) P(r)= a 3 2s/a 2 0 se ( ) ds P 4a = 0 0 0 4 0 a 2 3 2s/a 0 se ds. Using ( * ) from part (a) 0 (with b= 2/a ), 0 4a 2s/a 4 ( )= P 4a 0 a 0 e 3 8/a 0 (e) = 3 bx ( rp(r)dr= x e dx= e bx b = 4 a 3 4 3 2 4 8 ) 8 2 =1 41e 3 r e lim 4 s+2 a 0 0 t 3 a t 0 2 s+ a 0 1 82e 2 = 4 0 = 4 a 0 3 a 3 0 8 8 e (64+16+2) 1(2) 0 0.986 2r/a 0 dr. Integrating by parts three times or using a CAS, we find that 0 ( b3x3 3b2x2+6bx 6). So with b= 2