You take an exam that contains 20 multiple-choice questions. Each question has 4 possible options. You know the answer to 10 questions, but you have no idea about the other 10 questions so you choose answers randomly. Your score X on the exam is the total number of correct answers. Find: Range(X), the numbers P(X = 15), P(X ≥ 18), as well as E(X).

a) Range (X) = 10

b) P(X=15) = 0.0584

c) P(X ≥ 18) = 0.000416

d) E(X) = 12.5

Step-by-step explanation:

There are 20 questions in all.

The answers to 10 questions are very sure.

The other 10 questions are not known.

Minimum number of questions one can get right out of 20 = 10

Maximum number of questions one can get right out of 20 = 20

Probability of getting one of the remaining 10 questions correctly = (1/4) = 0.25 (since there are 4 possible options per question)

a) Range (X) = (Maximum number of questions one can get right) - (Minimum number of questions one can get right)

Range (X) = 20 - 10 = 10.

b) P(X=15)

It is confident that 10 questions are correct, so, P(X=15) is basically the probability of getting 5 questions correctly out of the remaining 10. P(X=15) = P(Y=5)

Using Binomial distribution formula

Binomial distribution function is represented by

P(Y = y) = ⁿCᵧ pʸ qⁿ⁻ʸ

n = total number of sample spaces = number of questions to guess from = 10

y = Number of successes required = 5

p = probability of success = probability of getting a question right = 0.25

q = probability of failure = probability of NOT getting a question right = 1 - 0.25 = 0.75

P(Y=5) = ¹⁰C₅ (0.25)⁵ (0.75)¹⁰⁻⁵ = ¹⁰C₅ (0.25)⁵ (0.75)⁵ = 0.058399 = 0.0584

P(X=15) = P(Y=5) = 0.0584

c) P(X ≥ 18) = P(X=18) + P(X=19) + P(X=20)

P(X ≥ 18) = P(Y ≥ 8) = P(Y=8) + P(Y=9) + P(Y=10)

= 0.0003862381 + 0.00002861023 + 0.0000009537 = 0.0004158019 = 0.000416

d) E(X) = Σ xᵢpᵢ

xᵢ = each question

pᵢ = probability of getting the question right

E(X) = 10(1×1) + 10(1×0.25) = 10 + 2.5 = 12.5

Hope this Helps!!!