Chapter 6, Problem 45CE

An auditor for Health Maintenance Services of Georgia reports 40% of policyholders 55 years or older submit a claim during the year. Fifteen policyholders are randomly selected for company records.

1. a. How many of the policyholders would you expect to have filed a claim within the last year?
2. b. What is the probability that 10 of the selected policyholders submitted a claim last year?
3. c. What is the probability that 10 or more of the selected policyholders submitted a claim last year?
4. d. What is the probability that more than 10 of the selected policyholders submitted a claim last year?

To determine

Find the number of policyholders who are expected to have filed a claim within the last year.

Answer to Problem 45CE

The number of policyholders who are expected to have filed a claim within the last year is 6.

Explanation of Solution

The number of policyholders who are expected to have filed a claim within the last year is calculated as follows:

$\begin{array}{c}E\left(x\right)=0.40\times 15\\ =6\end{array}$

Therefore, the number of policyholders who are expected to have filed a claim within the last year is 6.

To determine

Calculate the probability that 10 of the selected policyholders submitted a claim last year.

Answer to Problem 45CE

The probability that 10 of the selected policyholders submitted a claim last year is 0.0245.

Explanation of Solution

The formula to find the binomial probability is as follows:

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

Here, n=15 and π=0.40.

The probability that 10 of the selected policyholders submitted a claim last year is calculated as follows:

$\begin{array}{c}P\left(10\right)=C{}_{10}{\left(0.40\right)}^{10}{\left(1-0.40\right)}^{\left(15-10\right)}\\ =\frac{15!}{10!\left(15-10\right)!}\times {0.40}^{10}\times {0.60}^5\\ =3,003\times 0.0001\times 0.0778\\ =0.0245\end{array}$

Therefore, the probability that 10 of the selected policyholders submitted a claim last year is 0.0245.

To determine

Calculate the probability that 10 or more of the selected policyholders submitted a claim last year.

Answer to Problem 45CE

The probability that 10 or more of the selected policyholders submitted a claim last year is 0.0338.

Explanation of Solution

The probability that 10 or more of the selected policyholders submitted a claim last year is calculated as follows:

$\begin{array}{c}P\left(x\ge 10\right)=\left\{\begin{array}{l}\left[C{}_{10}{\left(0.40\right)}^{10}{\left(1-0.40\right)}^{\left(15-10\right)}\right]+\left[C{}_{11}{\left(0.40\right)}^{11}{\left(1-0.40\right)}^{\left(15-11\right)}\right]+\\ \left[C{}_{12}{\left(0.40\right)}^{12}{\left(1-0.40\right)}^{\left(15-12\right)}\right]+\left[C{}_{13}{\left(0.40\right)}^{13}{\left(1-0.40\right)}^{\left(15-13\right)}\right]+\\ \left[C{}_{14}{\left(0.40\right)}^{14}{\left(1-0.40\right)}^{\left(15-14\right)}\right]+\left[C{}_{15}{\left(0.40\right)}^{15}{\left(1-0.40\right)}^{\left(15-15\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\frac{15!}{10!\left(15-10\right)!}\times {0.40}^{10}\times {0.60}^5\right]+\left[\frac{15!}{11!\left(15-11\right)!}\times {0.40}^{11}\times {0.60}^4\right]+\\ \left[\frac{15!}{12!\left(15-12\right)!}\times {0.40}^{12}\times {0.60}^3\right]+\left[\frac{15!}{13!\left(15-13\right)!}\times {0.40}^{13}\times {0.60}^2\right]+\\ \left[\frac{15!}{14!\left(15-14\right)!}\times {0.40}^{14}\times {0.60}^1\right]+\left[\frac{15!}{15!\left(15-15\right)!}\times {0.40}^{15}\times {0.60}^0\right]\end{array}\right\}\\ =0.0245+0.0074+0.0016+0.0000+0.0000\\ =0.0338\end{array}$

Therefore, the probability that 10 or more of the selected policyholders submitted a claim last year is 0.0338.

To determine

Calculate the probability that more than 10 of the selected policyholders submitted a claim last year.

Answer to Problem 45CE

The probability that more than 10 of the selected policyholders submitted a claim last year is 0.0093.

Explanation of Solution

The probability that more than 10 of the selected policyholders submitted a claim last year is calculated as follows:

$\begin{array}{c}P\left(x>10\right)=\left\{\begin{array}{l}\left[C{}_{11}{\left(0.40\right)}^{11}{\left(1-0.40\right)}^{\left(15-11\right)}\right]+\left[C{}_{12}{\left(0.40\right)}^{12}{\left(1-0.40\right)}^{\left(15-12\right)}\right]+\\ \left[C{}_{13}{\left(0.40\right)}^{13}{\left(1-0.40\right)}^{\left(15-13\right)}\right]+\left[C{}_{14}{\left(0.40\right)}^{14}{\left(1-0.40\right)}^{\left(15-14\right)}\right]+\\ \left[C{}_{15}{\left(0.40\right)}^{15}{\left(1-0.40\right)}^{\left(15-15\right)}\right]\end{array}\right\}\\ =\left\{\begin{array}{l}\left[\frac{15!}{11!\left(15-11\right)!}\times {0.40}^{11}\times {0.60}^4\right]+\left[\frac{15!}{12!\left(15-12\right)!}\times {0.40}^{12}\times {0.60}^3\right]+\\ \left[\frac{15!}{13!\left(15-13\right)!}\times {0.40}^{13}\times {0.60}^2\right]+\left[\frac{15!}{14!\left(15-14\right)!}\times {0.40}^{14}\times {0.60}^1\right]+\\ \left[\frac{15!}{15!\left(15-15\right)!}\times {0.40}^{15}\times {0.60}^0\right]\end{array}\right\}\\ =0.0074+0.0016+0.0000+0.0000\\ =0.0093\end{array}$

Therefore, the probability that more than 10 of the selected policyholders submitted a claim last year is 0.0093.