Question

Problem 2-23

Suppose that the national average for the math portion of the College Board's SAT is 547. The College Board periodically rescales the test scores such that the standard deviation is approximately 75. Answer the following questions using a bell-shaped distribution and the empirical rule for the math test scores. If required, round your answers to two decimal places.

(a) What percentage of students have an SAT math score greater than 622? %

(b) What percentage of students have an SAT math score greater than 697? %

(c) What percentage of students have an SAT math score between 472 and 547? %

(d) What is the z-score for student with an SAT math score of 625?

(e) What is the z-score for a student with an SAT math score of 415?

Answer 1

Supposing that the distribution of the students is normal, the mean of the distribution is given as $\dpi{80} \mu = 547$ and standard deviation is given as $\dpi{80} \sigma = 75$ .

(a) A property of normal distribution is $\dpi{80} P(\mu - \sigma < x < \mu &plus; \sigma) = 0.68$ or $\dpi{80} P(x < \mu - \sigma) &plus; P(\mu &plus; \sigma > x) = 1 - 0.68 = 0.32$ , and since the distribution is symmetric, we have $\dpi{80} P(\mu &plus; \sigma > x) = 0.32/2$ or $\dpi{80} P(\mu &plus; \sigma > x) = 0.16$ . Putting the values of mean and SD, we have $\dpi{80} P(547 &plus; 75 > x) = 0.16$ or $\dpi{80} P(622 > x) = 0.16$ . Hence, probability of students having score over 622 is 16%.

(b) Another property of normal distribution is $\dpi{80} P(\mu - 2\sigma < x < \mu &plus; 2\sigma) = 0.95$ or $\dpi{80} P(x < \mu - 2\sigma) &plus; P(\mu &plus; 2\sigma > x) = 1 - 0.95 = 0.05$ , and since the distribution is symmetric, we have $\dpi{80} P(\mu &plus; 2\sigma > x) = 0.05/2$ or $\dpi{80} P(\mu &plus; 2\sigma > x) = 0.025$ . Putting the values of mean and SD, we have $\dpi{80} P(547 &plus; 2*75 > x) = 0.025$ or $\dpi{80} P(697 > x) = 0.025$ . Hence, probability of students having score over 697 is 2.5%.

(c) We have $\dpi{80} P(\mu - \sigma < x < \mu &plus; \sigma) = 0.68$ , and as the distribution is symmetric (around the mean), we have $\dpi{80} P(\mu - \sigma < x < \mu) = 0.68/2$ or $\dpi{80} P(\mu - \sigma < x < \mu) = 0.34$ . Putting the values of mean and SD, we have $\dpi{80} P(547 - 75 < x < 547) = 0.34$ or $\dpi{80} P(472 < x < 547) = 0.34$ . Hence, the probability of students having score between 472 and 547 is 34%.

(d) The z score would be $\dpi{80} Z = \frac{x - \mu}{\sigma}$ or $\dpi{80} Z = \frac{625 - 547}{75}$ or $\dpi{80} Z = \frac{78}{75}$ or $\dpi{80} Z = 1.04$ .

(e) The z score would be $\dpi{80} Z = \frac{x - \mu}{\sigma}$ or $\dpi{80} Z = \frac{415 - 547}{75}$ or $\dpi{80} Z = \frac{-132}{75}$ or $\dpi{80} Z = -1.76$ .