Exercise 71

Question 8.102

71. The scores of high school seniors on the ACT college entrance examination in 2010 were roughly normal with mean $μ = 21.0$ and standard deviation $σ = 5.2$ .

What is the approximate probability that a single student randomly chosen from all those taking the test scores 26.2 or higher?
Now take an SRS of nine students who took the test. What are the mean and standard deviation of the sample mean score $\bar{x}$ of these nine students?
What is the approximate probability that the mean score $\bar{x}$ of these nine students is 26.2 or higher?

71.

(a) Because 26.2 is 1 standard deviation above the mean, the probability is about 0.16.

(b) The mean remains $μ = 21.0$ . The standard deviation is $\frac{σ}{\sqrt{9}} = \frac{5.2}{3} = \frac{26}{15} \approx 1.7$ .

Page A-24

(c) Because $26.2 = 21.0 + 5.2 = 21.0 + 3 (\frac{26}{15})$ is 3 standard deviations above the mean, the probability is about 0.0015. (This is half of the 0.003 probability for outcomes more than 3 standard deviations from the mean, using the 99.7 part of the 68−95−99.7 rule.)