Chapter 5.3, Problem 26P

a.

To determine

Calculate the probability that all six professors are extroverts.

Answer to Problem 26P

The probability that all six professors are extroverts is 0.008.

Explanation of Solution

Calculation:

Let r follows a binomial distribution that represents the number of professors who are extroverts with success described as “extroverts” and failure described as “not extroverts”.

Total number of professors (or trials) $n=6$.

Probability of extrovert professors $p=0.45$.

Probability of non-extrovert professors (or failure) is calculated as given below:

$\begin{array}{c}q=1-0.45\\ =0.55\end{array}$

Binomial probability:

The probability of r successes out of n trials is given below:

$P(r)=C_{n,r}{p}^r{q}^{n-r}$

Here, n is the number of trials, r is the number of successes, p is the probability of success, and q is the probability of failure.

The probability that all six professors are extroverts is calculated as given below:

$\begin{array}{c}P\left(6\right)=C_{6,6}{\left(0.45\right)}^6{\left(0.55\right)}^0\\ =0.008\end{array}$

Thus, the probability that all six professors are extroverts is 0.008.

b.

To determine

Calculate the probability that none of the professors are extroverts.

Answer to Problem 26P

The probability that none of the professors are extroverts is 0.028.

Explanation of Solution

Calculation:

The probability that none of the professors are extroverts is calculated as given below:

$\begin{array}{c}P\left(0\right)=C_{6,0}{\left(0.45\right)}^0{\left(0.55\right)}^6\\ =0.028\end{array}$

Thus, the probability that none of the professors are extroverts is 0.028.

b.

To determine

Calculate the probability that none of the professors are extroverts.

Answer to Problem 26P

The probability that none of the professors are extroverts is 0.028.

Explanation of Solution

Calculation:

The probability that none of the professors are extroverts is calculated as given below:

$\begin{array}{c}P\left(0\right)=C_{6,0}{\left(0.45\right)}^0{\left(0.55\right)}^6\\ =0.028\end{array}$

Thus, the probability that none of the professors are extroverts is 0.028.

c.

To determine

Calculate the probability that at least two of the professors are extroverts.

Answer to Problem 26P

The probability that at least two of the professors are extroverts is 0.836.

Explanation of Solution

Calculation:

The probability that at least two of the professors are extroverts is calculated as given below:

$\begin{array}{c}P\left(r\ge 2\right)=P\left(2\right)+P\left(3\right)+P\left(4\right)+P\left(5\right)+P\left(6\right)\\ =\left\{\begin{array}{c}{C}_{6,2}{\left(0.45\right)}^2{\left(0.55\right)}^{\left(6-2\right)}+C_{6,3}{\left(0.45\right)}^3{\left(0.55\right)}^{\left(6-3\right)}+C_{6,4}{\left(0.45\right)}^4{\left(0.55\right)}^{\left(6-4\right)}+\\ C_{6,5}{\left(0.45\right)}^5{\left(0.55\right)}^{\left(6-5\right)}+C_{6,6}{\left(0.45\right)}^6{\left(0.55\right)}^{\left(6-6\right)}\end{array}\right\}\\ =0.278+0.303+0.186+0.061+0.008\\ =0.836\end{array}$

Thus, the probability that at least two of the professors are extroverts is 0.836.

d.

To determine

Calculate the expected number of extroverts.

Calculate the standard deviation.

Answer to Problem 26P

The expected number of extroverts is 2.7.

The standard deviation is 1.219.

Explanation of Solution

Calculation:

The expected number of extroverts is calculated as given below:

$\begin{array}{c}\mu =np\\ =6\times 0.45\\ =2.7\end{array}$

Thus, the expected number of extroverts is 2.7.

The standard deviation is calculated as given below:

$\begin{array}{c}\sigma =\sqrt{npq}\\ =\sqrt{np\left(1-p\right)}\\ =\sqrt{6\times 0.45\times \left(1-0.45\right)}\\ =1.219\end{array}$

Thus, the standard deviation is 1.219.

e.

To determine

Calculate the number of professors to be interviewed to be 90% certain that at least 3 professors will be extroverts.

Answer to Problem 26P

The number of professors to be interviewed to be 90% certain that at least 3 professors will be extroverts is 10.

Explanation of Solution

Calculation:

Compute n such that $P\left(r\ge 3\right)=0.90$.

The probability that at least 3 professors will be extroverts is calculated as given below:

$\begin{array}{l}P\left(r\ge 3\right)=1-\left[P\left(0\right)+P\left(1\right)+P\left(2\right)\right]\\ \text{ =}1-\left[0.003+0.021+0.076\right]\\ =1-0.100\\ (u\sin gbinomialdistributiontable)\\ =0.900\end{array}$

Using Table 3, Appendix II: Binomial distribution table, the probability values for $n=10$ and $p=0.45$ for $r=0,1,2$ are 0.003, 0.021 and 0.076.

Therefore, for $n=10$ the probability that at least 3 professors will be extroverts is 0.900.

Thus, 10 professors have to be interviewed to be 90% certain that at least 3 professors will be extroverts.