Question

1. (6 points) Isabel Briggs Myers was a pioneer in the study of personality types. Her work has been used successfully in counseling, educational, and industrial settings. In the book A Guide to the Development and Use of the Myers-Briggs Type Indicators, by Myers and McCaully, it was reported that based on a very large sample (2282 professors), approximately 45% of all university professors are extroverted.

Suppose a random sample of eight professors is selected from this survey. Answer the following questions. Include your formula setup or calculator comments for each question.

1. (1 pt) What is the probability that all eight are extroverts?

2. (1 pt) What is the probability that none of the professors in the sample is an extrovert?

3. (1 pt) What is the probability that at least three of the professors in the sample are extroverts?

4. (1 pt) What is the probability that at most four professors in this sample are extroverts?

5. (2 pts) In this group of eight professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution? Show your formula for each

Answer 1

p = 0.45

n = 8

a) P(X = 8) = 0.45⁸ = 0.0017

b) P(X = 0) = (1 - 0.45)⁸ = 0.0084

c) P(X > 3) = 1 - (P(X = 0) + P(X = 1) + P(X = 2))

                  = 1 - (8C0 * 0.45⁰ * 0.55⁸ + 8C1 * 0.45¹ * 0.55⁷ + 8C2 * 0.45² * 0.55⁶)

                  = 1 - 0.2201

                  = 0.7799

d) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

                  = 8C0 * 0.45⁰ * 0.55⁸ + 8C1 * 0.45¹ * 0.55⁷ + 8C2 * 0.45² * 0.55⁶ + 8C3 * 0.45³ * 0.55⁵ + 8C4 * 0.45⁴ * 0.55⁴

                  = 0.7396

e) Expected number = n * p = 8 * 0.45 = 3.6

Standard deviation = sqrt(n * p * (1 - p)) = sqrt(8 * 0.45 * 0.55) = 1.41