A student takes a multiple-choice test with 40 questions. The probability that the student answers a given question correctly is 0.5, independent of all other questions. The probability that the student answers more than [math]N[/math] questions correctly is greater than 0.10. The probability that the student answers more than [math]N+1[/math] questions correctly is less than 0.10.

Calculate [math]N[/math] using a normal approximation with the continuity correction.

• 23
• 25
• 32
• 33
• 35

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: A

Let C be the number correct. C has a binomial distribution with n = 40 and p = 0.5. Then the mean is 40(0.5) = 20 and the variance is 40(0.5)(0.5) = 10. With the exact probability we have

[[math]] \begin{align*} 0.1 = \operatorname{P}(C \gt N) = \operatorname{P}(Z \gt \frac{N + 0.5 − 20 }{\sqrt{10}}) \\ 1.282 = \frac{N+0.5 -20}{\sqrt{10}}, \, N = 1.282 \sqrt{10} + 19.5 = 23.55. \end{align*} [[/math]]

At [math]N=23[/math] the approximate probability will exceed 0.1 ([math]Z=1.107[/math]).

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.