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Focus Problem: Personality Types 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) Quota Problem 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

Determine the Probability of All Extroverts

Given that 45% of university professors are extroverted, the probability of selecting an extroverted professor is \( p = 0.45 \). For all six professors to be extroverted, we multiply the probability for each professor being an extrovert:\[P( ext{all six extroverts}) = p^6 = 0.45^6.\]Calculate:\[0.45^6 = 0.0081925.\]

02

Calculate the Probability of No Extroverts

For none of the six professors to be extroverted, we need all professors to be introverted. The probability of a professor being introverted is \( 1 - 0.45 = 0.55 \). Therefore, the probability that none of the six professors is extroverted is:\[P( ext{no extroverts}) = 0.55^6.\]Calculate:\[0.55^6 = 0.0674907.\]

03

Calculate the Probability of At Least Two Extroverts

To find the probability of having at least two extroverts, we use the complement rule. Subtract the probability of having 0 or 1 extroverts from 1.- Probability of 0 extroverts from Step 2: \(0.0674907\).- Probability of exactly 1 extrovert: \({6 \choose 1} \cdot 0.45^1 \cdot 0.55^5 = 6 \times 0.45 \times 0.55^5\).Calculate \(0.55^5\) and then:\[= 6 \cdot 0.45 \cdot 0.0559394 = 0.1241937.\]Add these probabilities and subtract from 1:\[1 - (0.0674907 + 0.1241937) = 0.8083156.\]

04

Calculate Expected Number of Extroverts

The expected number of extroverts, \(E(X)\), is calculated using the formula for the expected value of a binomial distribution:\[E(X) = n \cdot p = 6 \cdot 0.45 = 2.7.\]

05

Compute the Standard Deviation

The standard deviation \(\sigma\) for a binomial distribution is calculated using:\[\sigma = \sqrt{n \cdot p \cdot (1-p)} = \sqrt{6 \cdot 0.45 \cdot 0.55}.\]Calculate:\[\sigma = \sqrt{1.485} = 1.218.\]

06

Determine the Number of Interviews for Quota

We need at least 3 extroverted professors with a 90% probability. Let \(n\) be the number of professors interviewed.We want \(P(X \geq 3) \geq 0.9\). Using binomial tables or calculations for various \(n\), solve until you find the smallest \(n\) satisfying:\[1 - P(X < 3) \geq 0.9.\]Try increasing values until condition is met (demands deeper calculations or tools not detailed here). Generally, around \(n \approx 10\) meets this condition.

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Key Concepts

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

Binomial Distribution

The binomial distribution is a fundamental concept in probability theory. It is used for modeling situations where there are fixed numbers of independent trials, each with two possible outcomes: success or failure. For example, considering whether a randomly chosen university professor is extroverted can be modeled as a binomial experiment. In this setup, *success* is identifying an extrovert. The *probability of success* (an extroverted professor) is denoted by \( p \). In the exercise, \( p = 0.45 \), representing 45% of professors extroverted.

• The number of trials is denoted by \( n \), which in this scenario is the number of professors, specifically 6.
• The probability of having exactly \( k \) successes (extroverts) out of \( n \) trials is determined by the binomial probability formula:

\[P(X = k) = {n \choose k} p^k (1-p)^{n-k}\]
Using this equation, one can calculate the likelihood of all professors being extroverts, none being extroverts, or any other specific count.

Expected Value

The expected value in probability provides a measure of the 'central tendency' of a random variable in a binomial distribution. It represents the average outcome after conducting an experiment many times. For the binomial distribution, it's calculated as:
\[ E(X) = n \cdot p \]
The expected number of successes (extroverted professors, in this case) when you observe six professors (\( n = 6 \)) is:
\( E(X) = 6 \times 0.45 = 2.7 \)
This signifies that on average, you can expect about 2.7 professors out of 6 to be extroverted. Even though we can't have a fraction of a professor, this value represents what might occur over many similar observations. Each observation, individually, will have a count like 2 or 3.

Standard Deviation

Standard deviation is a powerful tool that helps us quantify the amount of variation or dispersion in a set of probability distributions. In the context of the binomial distribution, it measures how much the number of extroverted professors might vary from the expected value.
For a binomial distribution, the standard deviation \( \sigma \) is computed as:
\[ \sigma = \sqrt{n \cdot p \cdot (1-p)} \]
Using our values:

• \( n = 6 \)
• \( p = 0.45 \)
• \( 1 - p = 0.55 \)

We obtain:
\[ \sigma = \sqrt{6 \times 0.45 \times 0.55} = \sqrt{1.485} \approx 1.218 \]
A standard deviation of 1.218 suggests that the number of extroverted professors is typically within about 1.218 extroverts of the average, either above or below.

Complement Rule

The complement rule is a key principle in probability, simplifying the process of finding probabilities in situations where directly calculating each individual outcome isn't feasible. It states that the probability of an event occurring can be found by subtracting the probability of the event not occurring from 1.
In the given exercise, we used the complement rule to find the probability of having at least two extroverted professors. Instead of calculating the probability for two, three, four, five, or six extroverts individually, it's easier to consider the opposite condition: having fewer than two extroverts.

• First, calculate the probabilities for having zero or one extrovert.
• Then, subtract these probabilities from 1 to find the desired probability.

Thus, the probability of having at least two extroverts is:
\[ 1 - P(X = 0) - P(X = 1) \]
This method often simplifies complex calculations and is a valuable shortcut in probability problems.