Answer to Question #224304 in Statistics and Probability for ali

Question

A multiple choice quiz has 15 questions, each with 4 possible answers, of which one is correct. How many students would you expect to get more than 10 marks if a class contains 100 students?

Answer 1

Given that there are 4 possible choices.

Suppose a student guesses on each question.

The probability of getting the correct answer is $= \frac{1}{4}=0.25$

X=the number of correct questions

n=number of trials=15

p=probability of success=0.25

X follows a binomial distribution with n=10 and p=0.25

$P(X=x) = C^n_xp^x(1-p)^{n-x} \\ P(X>10) = P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15) \\ P(X=11) = \frac{15!}{11!(15-11)!} \times 0.25^{11} \times 0.75^{15-11} \\ = 0.000102 \\ P(X=12) = \frac{15!}{12!(15-12)!} \times 0.25^{12} \times 0.75^{15-12} \\ = 1.14 \times 10^{-5} \\ P(X=13) = \frac{15!}{13!(15-13)!} \times 0.25^{13} \times 0.75^{15-13} \\ = 8.80 \times 10^{-7} \\ P(X=14) = \frac{15!}{14!(15-14)!} \times 0.25^{14} \times 0.75^{15-14} \\ = 4.19 \times 10^{-8} \\ P(X=15) = \frac{15!}{15!(15-15)!} \times 0.25^{15} \times 0.75^{0} \\ = 9.31 \times 10^{-10} \\ P(X>10) = 0.000102 + (1.14 \times 10^{-5}) + (8.80 \times 10^{-7}) +(4.19 \times 10^{-8}) + (9.31 \times 10^{-10}) \\ ≈0.0001$

N=100

Number of students with than 10 marks $= 100 \times 0.0001 = 0.01 ≈ 0$