Question

Assume that on a standardized test of 100 questions, a person has a probability of 80% of answering any particular question correctly. Find the probability of answering between 70 and 80 questions, inclusive. (Assume independence, and round your answer to four decimal places.)

P(70 ? X ? 80) =

Answer 1

P(answering a question correctly), p = 0.80

q = 1 - p = 0.20

Sample size, n = 100

Normal approximation for binomial distribution: P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 100 x 0.80

= 80

Standard deviation = $\sqrt{npq}$

= $\sqrt{100\times0.8\times0.2}$

= 4

P(answering between 70 and 80 questions, inclusive) = P(70 $\leq$ X $\leq$ 80)

= P(X < 80.5) - P(X < 69.5) (with continuity correction)

= P(Z < (80.5 - 80)/4) - P(Z < (69.5 - 80)/4)

= P(Z < 0.125) - P(Z < -2.625)

= 0.5497 - 0.0043

= 0.5454