Question

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 pt) What is the probability that all eight are extroverts?
(1 pt) What is the probability that none of the professors in the sample is an extrovert?
(1 pt) What is the probability that at least two of the professors in the sample are extroverts?
(1 pt) What is the probability that at most four professors in this sample are extroverts?
(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

a)
`P(X=8)=(8C8)(0.45)^8 (1-0.45)^(8-8)=0.00168`

b)
`P(X=0)=(8C0)(0.45)^0 (1-0.45)^(8-0)=0.00837`

c)
`P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=0.7396`

d)
`E(X) = np = 8*0.45= 3.6`

The variance is given by:

`Var(X) =np(1-p)= 8*0.45*(1-0.45)=1.98`

And the deviation would be:

`Sd(X) = √(1.98)= 1.407`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Part a

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=8, p=0.45)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

We want to find this probability :
`P(X=8)`

Using the pmf we have:

`P(X=8)=(8C8)(0.45)^8 (1-0.45)^(8-8)=0.00168`

Part b

We want to find this probability :
`P(X=0)`

Using the probability mass function we got:

`P(X=0)=(8C0)(0.45)^0 (1-0.45)^(8-0)=0.00837`

Part c

We want this probability:

`P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)`

We find the individual probabilities like this:

`P(X=0)=(8C0)(0.45)^0 (1-0.45)^(8-0)=0.00837`

`P(X=1)=(8C1)(0.45)^1 (1-0.45)^(8-1)=0.0548`

`P(X=2)=(8C2)(0.45)^2 (1-0.45)^(8-2)=0.1569`

`P(X=3)=(8C3)(0.45)^3 (1-0.45)^(8-3)=0.2568`

`P(X=4)=(8C4)(0.45)^4 (1-0.45)^(8-4)=0.2627`

And adding the values we got:

`P(X \leq 4) = P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)=0.7396`

Part d

The expected value is given by:

`E(X) = np = 8*0.45= 3.6`

The variance is given by:

`Var(X) =np(1-p)= 8*0.45*(1-0.45)=1.98`

And the deviation would be:

`Sd(X) = √(1.98)= 1.407`