Section 9.1 Exercise 3

A true-false examination has 48 questions. June has probability 3/4 of an- swering a question correctly. April just guesses on each question. A passing score is 30 or more correct answers. Compare the probability that June passes the exam with the probability that April passes it.

This problem can be model using the binomial distribution and approximated using the Central Limit Theorem.

\[P(i \le S_n \le j) \approx NA(\frac{i-1/2-np}{\sqrt{npq}}, \frac{j +1/2-np}{\sqrt{npq}})\]

For June:

\[P(S_{48} \ge 30) = P(S^*_{48} \ge \frac{30-1/2-48*0.75}{\sqrt{48*0.75*0.25}}) = P(S^*_{48} \ge - 2.166)\] First we need to recognize that the table of values of NA(0,z) in Figure 9.4 takes z values from 0 to z. So we need to rewrite our problem as the below and interpolate between z values of NA(2.1) = 0.4821 and NA(2.2) = 0.4861 .

\[P(S_{48} \ge 30) = 0.5 + NA(2.166)\] \[P(S_{48} \ge 30) \approx 0.5 + 0.4821 + (0.4861-0.4821)/10*6 \approx 0.9845\]

For April:

\[P(S_{48} \ge 30) = P(S^*_{48} \ge \frac{30-1/2-48*0.5}{\sqrt{48*0.5*0.5}}) = P(S^*_{48} \ge 1.5877)\]

We must interpolate between NA(1.5) = 0.4332 and NA(1.6) = 0.4452. This time we are interested in the area under the curve from the z value to the upper tail. So we must subtract the NA(z) value from 0.5 to get the area of interest.

\[P(S_{48} \ge 30) = 0.5 - NA(1.5877)\]

\[P(S_{48} \ge 30) \approx 0.5 - (0.4332 + (0.4452-0.4332)/10*8.7) \approx 0.05636\]