ch13 - Chapter 13 Solutions 13.1. Yes: (1) We have a fixed...

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Chapter 13 Solutions 13.1. Yes: (1) We have a fxed number oF observations ( n = 15). (2) It is reasonable to believe that each call is independent oF the others. (3) “Success” means reaching a live person, “Failure” is any other outcome. (4) Each randomly dialed number has chance p = 0 . 2o F reaching a live person. 13.2. Not binomial: we do not have a fxed number oF observations. 13.3. Not binomial: because the student receives instruction aFter incorrect answers, her probability oF success is likely to increase. 13.4. The number in the sample who say they Feel pressure to go to a good college has approximately the binomial distribution with n = 1000 and p = 0 . 63. 13.5. (a) C , the number caught, is binomial with n = 10 and p = 0 . 7. M , the number missed, is binomial with n = 10 and p = 0 . 3. (b) We fnd P ( M = 3 ) = ( 10 3 ) ( 0 . 3 ) 3 ( 0 . 7 ) 7 = 120 ( 0 . 027 )( 0 . 08235 ) . = 0 . 2668. With soFtware, we fnd P ( M 3 ) . = 0 . 6172. 13.6. Let N be the number oF live persons contacted among the 15 calls observed. Then N has the binomial distribution with n = 15 and p = 0 . 2 (see also the solution to Exercise 13.1). (a) P ( N = 3 ) = ( 15 3 ) ( 0 . 2 ) 3 ( 0 . 8 ) 12 . = 0 . 2501. (b) P ( N 3 ) = P ( N = 0 ) +···+ P ( N = 3 ) . = 0 . 6482. (c) P ( N 3 ) = P ( N = 3 ) P ( N = 15 ) . = 0 . 6020. (d) P ( N < 3 ) = P ( N = 0 ) P ( N = 2 ) . = 0 . 3980. (e) P ( N > 3 ) = P ( N = 4 ) P ( N = 15 ) . = 0 . 3518. The TI-83 screen on the right illustrates the use oF that calculator’s binompdf and binomcdf Functions (Found in the DISTR menu) to compute the frst two probabilities. The frst oF these fnds individual binomial probabilities, and the second fnds cumulative prob- abilities (that is, it sums the probability From 0 up to a given number). Excel oFFers similar Features with its BINOMDIST Function. Calculators that do not have binomial probabilities may have a built-in Function to compute, e.g., ( 15 3 ) . 163
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164 Chapter 13 Binomial Distributions* 13.7. The screenshots below show Google’s answers at the time these solutions were prepared. (a) 5 choose 2 returns 10. (b) 500 choose 2 returns 124,750, and 500 choose 100 returns 2 . 04169424 × 10 107 . (c) (10 choose 1)*0.1*0.9^9 returns 0.387420489. 13.8. (a) With n = 15 and p = 0 . 2, we have µ = np = 3 calls. (b) σ = ( 1 p ) = 2 . 4 . = 1 . 5492 calls. (c) With p = 0 . 08, σ = 1 . 104 . = 1 . 0507 calls; with p = 0 . 01, σ = 0 . 1485 . = 0 . 3854 calls. As p approaches 0, the standard deviation decreases (that is, it also approaches 0). 13.9. (a) X is binomial with n = 10 and p = 0 . 3; Y is binomial with n = 10 and p = 0 . 7 (b) The mean of Y is ( 10 )( 0 . 7 ) = 7 errors caught, and for X the mean is ( 10 )( 0 . 3 ) = 3 errors missed. (c) The standard deviation of Y (or X )is σ = ( 10 )( 0 . 7 )( 0 . 3 ) . = 1 . 4491 errors. 13.10. Let X be the number of 1’s and 2’s; then X has a binomial distribution with n = 90 and p = 0 . 477 (in the absence of fraud). This should have a mean of 42.93 and standard deviation σ . = 22 . 4524 = 4 . 7384. Therefore, P ( X 29 ) . = P ³ Z 29 42 . 93 4 . 7384 ´ = P ( Z ≤− 2 . 94 ) = 0 . 0016 . (Using software, we Fnd that the exact value is 0.0021.) This probability is quite small, so we have reason to be suspicious. 13.11. (a) µ = ( 1535 )( 0 . 27 ) = 414 . 45 and σ = 302 . 549 .
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This note was uploaded on 11/23/2010 for the course STAT 2325151 taught by Professor T during the Spring '10 term at Waters College.

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ch13 - Chapter 13 Solutions 13.1. Yes: (1) We have a fixed...

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