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We now have the tools to solve the chapter Focus Problem. In the book \(A\) Guide to the Development and Use of the Myers-Briggs Type Indicators by Myers and McCaully, it was reported that approximately \(45 \%\) of all university professors are extroverted. Suppose you have classes with six different professors. (a) What is the probability that all six are extroverts? (b) What is the probability that none of your professors is an extrovert? (c) What is the probability that at least two of your professors are extroverts? (d) In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution? (e) Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least \(90 \%\) sure of filling the quota?

Step by step solution

01

Identify Binomial Distribution

The problem deals with a binomial distribution because we have a fixed number of professors (six), and each professor can either be an extrovert or not. The probability of being extroverted is given as \( p = 0.45 \). The number of trials is \( n = 6 \).

02

Probability of All Professors Being Extroverts (Part a)

Use the binomial probability formula, \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \). For all six professors being extroverts (\( k = 6 \)), calculate \( P(X=6) = \binom{6}{6} (0.45)^6 (0.55)^{0} = 0.45^6 \). Evaluating this gives approximately 0.00878.

03

Probability of No Professors Being Extroverts (Part b)

For none of the professors being extroverts (\( k = 0 \)), calculate \( P(X=0) = \binom{6}{0} (0.45)^0 (0.55)^6 = 0.55^6 \). Evaluating this gives approximately 0.07776.

04

Probability of At Least Two Professors Being Extroverts (Part c)

Calculate \( P(X \geq 2) \) using the complement rule: \( P(X \geq 2) = 1 - P(X < 2) = 1 - (P(X=0) + P(X=1)) \). We've already calculated \( P(X=0) = 0.07776 \), now calculate \( P(X=1) = \binom{6}{1} (0.45)^1 (0.55)^5 \), then add \( P(X=0) + P(X=1) \) and subtract from 1. This yields \( 0.382 \) for \( P(X \geq 2) \).

05

Expected Number of Extroverts and Standard Deviation (Part d)

For a binomial distribution, the expected value \( E(X) = n \times p \). Thus, \( E(X) = 6 \times 0.45 = 2.7 \). The standard deviation \( \sigma = \sqrt{n \times p \times (1-p)} = \sqrt{6 \times 0.45 \times 0.55} \), resulting in approximately \( 1.21 \).

06

Minimum Interviews to Meet Quota (Part e)

Use cumulative binomial probabilities to find the smallest \( n \) such that \( P(X \geq 3) \geq 0.90 \). This involves trial and error or using binomial tables, incrementing \( n \) until the accumulated probability from \( \binom{n}{3} \) to \( \binom{n}{n} \) sums to at least 0.90. The calculations will show that you need this for \( n = 8 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability

Understanding probability is crucial when dealing with binomial distributions. Probability essentially measures the likelihood of an event occurring.
In this context, we have a scenario where each professor can either be an extrovert or not. The probability of each professor being an extrovert is given as 0.45. This means that for each professor, there's a 45% chance they will be an extrovert.

• In part (a) of the exercise, we calculate the probability that all six professors are extroverts. We use the binomial formula since this is a binomial distribution problem, which deals with a fixed number of independent trials (in this case, six professors), each having two possible outcomes (extroverted or not).
  
• For part (b), we calculate the probability of none being an extrovert, showcasing the concept of complementary probability.
• Part (c) uses the complement rule which helps in finding probabilities for compounded events, specifically here, in finding the chance that at least two professors are extroverts by subtracting the probabilities of fewer than two. This exercise helps illustrate that probabilities of multiple events can often be calculated efficiently with complements.

Expectation in Statistics

Expectation in statistics, especially in the context of probability distributions, is a vital concept. It is sometimes called the expected value and it provides a measure of the center or "average" in probability terms. For a binomial distribution, the expected value is found by multiplying the total number of trials by the probability of success in each trial.
Hence, in part (d) of the exercise, the expected number of extroverted professors is calculated as follows:

• Given six professors (\( n = 6 \)) and a 45% probability of extroversion (\( p = 0.45 \)), the expected number of extroverts, \( E(X) = n \times p \ = 6 \times 0.45 = 2.7 \).

This expectation tells us that out of six randomly chosen professors, we can expect that about 2.7 of them to be extroverted on average. This doesn’t mean you can have 0.7 of a person; rather, it reflects the average across many such groups of six professors. It helps to set realistic expectations about outcomes when analyzing random events.

Standard Deviation

Standard deviation offers insight into the variability or dispersion of probabilities within a distribution. For binomial distributions, the formula to find the standard deviation (\( \sigma \)) is a bit more complex than the straightforward mean.
The formula involves:

• Calculating the standard deviation as \( \sigma = \sqrt{n \times p \times (1-p)} \).
• In the exercise, with \( n = 6 \) and \( p = 0.45 \), the standard deviation is \( \sigma = \sqrt{6 \times 0.45 \times 0.55} \), which evaluates to approximately 1.21.

This figure indicates how much the number of extroverted professors might vary from the expected number of 2.7 when you randomly sample groups of six professors. A smaller standard deviation would mean the numbers of extroverts per group are closely clustered around the mean, indicating less variability, whereas a larger standard deviation would suggest a wider spread of these numbers. Understanding standard deviation is crucial because it provides depth to interpreting statistical data, illustrating how much variability there is around a central expectation.