Question

1. A test has 30 questions, where each question has a choice of four answers, only one of which is correct. If a student answers the questions by randomly selecting answers, what is the probability that they will get 8 or more questions correct?

Answer 1

P(getting correct answer), p = 1 / Number of choices

= 1/4

= 0.25

q = 1 - p = 0.75

n = 30

np = 30x0.25 = 7.5

nq = 30x0.75 = 22.5

np and nq > 5

n $\geq$ 30

So, according to central limit theorem, normal approximation can be used. P(X < A) = P(Z < (A - mean)/standard deviation)

Normal approximation for binomial distribution: Mean = np

= 7.5

Standard deviation = $\sqrt{npq}$

= $\sqrt{30\times 0.25\times 0.75}$

= 2.372

P(a student will get 8 or more correct) = P(X $\geq$ 8)

= 1 - P(X < 7.5) (continuity correction of 0.5 is applied)

= 1 - P(Z < (7.5 - 7.5)/2.372)

= 1 - P(Z < 0)

= 1 - 0.5 (from standard normal distribution table)

= 0.5