Question

Suppose two friends, Michael and Austin, are arguing about Michael's supposed trivia prowess. Michael, a trivia fanatic, wishes to prove his vast trivia knowledge by answering a series of trivia questions. To test this, Austin purchases a book of trivia questions and randomly selects 100 to ask Michael. Assume that for each question, the probability that Michael gives the correct answer is 0.85.

What is the probability that Michael will answer at least 80 questions correctly? Use the normal approximation to obtain the result. Give your answer as a decimal precise to at least three decimal places.

?(?≥80)=

Answer 1

Since we are using the normal approximation to the binomial, we need to use the continuity correction factors, wherever applicable). We then find the Z scores, and use the standard normal to find the probabilities or use the EXCEL formula NORMSDIST(Z value).

Z = (X - )/

Continuity Correction Factor Table

1) If P(X = n)       Use P(n – 0.5 < X < n+0.5)

2) If P(X > n)       Use P(X > n+0.5)

3) If P(X <= n)    Use P(X < n+0.5)

4) If P(X < n)       Use P(X < n-0.5)

5) If P(X => n)    Use P(X > n+0.5)

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n = 100, p = 0.85 , q = 1 - p = 0.15

n*p = 100 * 0.85 = 85 and n * (1 - p) = 100 * 0.15 = 15. Since both are > 10, we can use the normal approximation to the binomial.

Mean = n * p = 85

Standard deviation, = Sqrt( n * p * (1 - p)) = sqrt(100 * 0.85 * 0.15) = 3.57

To Find P(X 80).

Using the correction Factor for P(X n) = P(X > n - 0.5)

Therefore P(X 80) = P(X > 80 - 0.5) = P(X > 79.5)

P(X > 79.5) = 1 - P(X < 79.5)

Z = (79.5 - 85)/3.57 = -1.541

The probability for (X < 79.5) = 0.0617

Therefore the required probability for P(X > 79.5) = 1 - 0.0617 = 0.9383 0.938 (Rounding to 3 decimal places)

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