Homework #6--Suggested Answers

ISBE 4.17, 4.19, 4.22, 4.29

4.17

1. Then, Pr(T > 15) = 0.5*(base)*(height) = 0.5*(5)*(0.05) = 0.125.
2. E[T] = 10. The distribution of T is both unimodal and symmetric. Therefore, the mean of T coincides with the mode of T.

4.19

1. Use table IV (inside front cover of the text). Pr(Z>1.6) = 0.055.
2. P(1.6<Z<2.3) = 0.055 - 0.011 = 0.044.
3. P(Z<1.64) = 1-0.051 = 0.949.
4. P(-1.64<Z<-1.02) = 0.154 - 0.051 = 0.103.
5. P(0<Z<1.96) = 0.5 - 0.025 = 0.475
6. P(-1.96<Z<1.96) = 0.475+0.475 = 0.95.
7. P(-1.5<Z<0.67) = (0.5-0.067) + (0.5-0.251) = 0.433 + 0.249 = 0.682.
8. P(Z<-2.5) = 0.006.

4.22

1. Pr(X>120) = Pr(Z>.5) = 0.309, thus the proportion of students finishing within 2 hours is 1-Pr(Z>.5) = 0.691, or 69.1%
2. 135.6 minutes. First, find the value of z such that Pr(Z>=z) = 0.1 by finding 0.1 in Table IV and identifying the associated z score, z=1.28. Use equation 4-23 to solve for X, which is 135.6.

4.29

Answers A, B, and E are true as they are linear transformations of the random variable X. In D, 10 does not equal 9. In C, 3.75 does not equal 10/3.