Chapter 6, Problem 23E

a.

To determine

Find the number of problems that is expected to be resolved today.

Find the standard deviation.

Answer to Problem 23E

The number of problems that is expected to be resolved today is 10.5.

The standard deviation is 1.7748.

Explanation of Solution

Here, n=15; π=0.70.

The expected number of problems to be resolved today is calculated as follows:

$\begin{array}{c}\mu =n\pi \\ =15\times 0.7\\ =10.5\end{array}$

Therefore, the expected number of problems to be resolved today is 10.5.

The standard deviation is calculated as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{15\times 0.7\times \left(1-0.7\right)}\\ =\sqrt{3.15}\\ =1.7748\end{array}$

Therefore, the standard deviation is 1.7748.

b.

To determine

Compute the probability that 10 of the problems can be resolved today.

Answer to Problem 23E

The probability that 10 of the problems can be resolved today is 0.2061.

Explanation of Solution

The formula to find the binomial probability is as follows:

$\begin{array}{l}P\left(X\right)=C{}_x{\pi }^x{\left(1-\pi \right)}^{\left(n-x\right)}\\ \text{where, }C\text{ is the combination}\text{.}\\ n\text{ is the number of trials}\text{.}\\ X\text{ is the random variable}\text{.}\\ \pi \text{ is the probability of success}\text{.}\end{array}$

The probability that 10 of the problems can be resolved today is calculated as follows:

$\begin{array}{c}P\left(x=10\right)=C{}_{10}{\left(0.70\right)}^{10}{\left(1-0.70\right)}^{\left(15-10\right)}\\ =\frac{15!}{10!\left(15-10\right)!}\times {0.70}^{10}\times {0.30}^5\\ =3003\times 0.0282\times 0.0024\\ =0.2061\end{array}$

Therefore, the probability that 10 of the problems can be resolved today is 0.2061.

c.

To determine

Compute the probability that 10 or 11 of the problems can be resolved today.

Answer to Problem 23E

The probability that 10 or 11 of the problems can be resolved today is 0.4247.

Explanation of Solution

The probability that 10 or 11 of the problems can be resolved today is calculated as follows:

$\begin{array}{c}P\left(x=10\right)+P\left(x=11\right)=\left(C{}_{10}{\left(0.70\right)}^{10}{\left(1-0.70\right)}^{\left(15-10\right)}\right)+\left(C{}_{11}{\left(0.70\right)}^{11}{\left(1-0.70\right)}^{\left(15-11\right)}\right)\\ =\left(\frac{15!}{10!\left(15-10\right)!}\times {0.70}^{10}\times {0.30}^5\right)+\left(\frac{15!}{11!\left(15-11\right)!}\times {0.70}^{11}\times {0.30}^4\right)\\ =\left(3,003\times 0.0282\times 0.0024\right)+\left(1,365\times 0.0198\times 0.0081\right)\\ =0.2061+0.2186\\ =0.4247\end{array}$

Therefore, the probability that 10 or 11 of the problems can be resolved today is 0.4247.

d.

To determine

Compute the probability that more than 10 of the problems can be resolved today.

Answer to Problem 23E

The probability that more than 10 of the problems can be resolved today is 0.5154.

Explanation of Solution

The probability that more than 10 of the problems can be resolved today is calculated as follows:

$\begin{array}{c}P\left(x>10\right)=P\left(x=11\right)+P\left(x=12\right)+P\left(x=13\right)+P\left(x=14\right)+P\left(x=15\right)\\ =\left\{\begin{array}{c}\left(C{}_{11}{\left(0.70\right)}^{11}{\left(1-0.70\right)}^{\left(15-11\right)}\right)+\left(C{}_{12}{\left(0.70\right)}^{12}{\left(1-0.70\right)}^{\left(15-12\right)}\right)+\\ \left(C{}_{13}{\left(0.70\right)}^{13}{\left(1-0.70\right)}^{\left(15-13\right)}\right)+\left(C{}_{14}{\left(0.70\right)}^{14}{\left(1-0.70\right)}^{\left(15-14\right)}\right)+\\ \left(C{}_{15}{\left(0.70\right)}^{15}{\left(1-0.70\right)}^{\left(15-15\right)}\right)\end{array}\right\}\\ =\left[\begin{array}{l}\left(\frac{15!}{11!\left(15-11\right)!}\times {0.70}^{11}\times {0.30}^4\right)+\left(\frac{15!}{12!\left(15-12\right)!}\times {0.70}^{12}\times {0.30}^3\right)+\\ \left(\frac{15!}{13!\left(15-13\right)!}\times {0.70}^{13}\times {0.30}^2\right)+\left(\frac{15!}{14!\left(15-14\right)!}\times {0.70}^{14}\times {0.30}^1\right)+\\ \left(\frac{15!}{15!\left(15-15\right)!}\times {0.70}^{15}\times {0.30}^0\right)\end{array}\right]\\ =0.2186+0.1700+0.0916+0.0305+0.0047\\ =0.5154\end{array}$

Therefore, the probability that more than 10 of the problems can be resolved today is 0.5154.