Question

Twenty-five psychology instructors have formed a committee to pick next year's textbook, and they have narrowed their decision down to two equally good books, one with a better bibliography and references, and the other with a better format and illustrations. Since the books are considered to be equally good, we will assume the probability an instructor chooses either book is 0.5 and the instructors' decisions are made independently. Using the binomial distribution, find the probability 15 or more instructors choose the book with the better format and illustrations.

Answer 1

To find the probability that 15 or more instructors choose the book with the better format and illustrations, we can use the binomial distribution formula.

Let's denote the event of an instructor choosing the book with the better format and illustrations as "success" (S), and the event of an instructor choosing the other book as "failure" (F). The probability of success is p = 0.5, and the probability of failure is q = 1 - p = 0.5.

We want to find the probability of 15 or more successes in a sample of 25 instructors. We can calculate this probability by summing the probabilities of exactly 15, 16, 17, ..., 25 successes.

P(X ≥ 15) = P(X = 15) + P(X = 16) + ... + P(X = 25)

Using the binomial distribution formula, the probability of exactly k successes in a sample of n trials is:

P(X = k) = C(n, k) * p^k * q^(n-k)

where C(n, k) is the binomial coefficient "n choose k," given by:

C(n, k) = n! / (k! * (n-k)!)

Applying this to our problem, we can calculate the probability as follows:

P(X ≥ 15) = P(X = 15) + P(X = 16) + ... + P(X = 25)

= Σ[ k = 15 to 25 ] ( C(25, k) * p^k * q^(25-k) )

Let's calculate this probability using the binomial distribution formula:

P(X ≥ 15) = Σ[ k = 15 to 25 ] ( C(25, k) * (0.5)^k * (0.5)^(25-k) )

Calculating this sum gives us the probability that 15 or more instructors choose the book with the better format and illustrations.