Life insurance A life insurance company sells a term insurance policy to 21-year-old males that pays \(100,000 if the insured dies within the next 5 years. The probability that a randomly chosen male will die each year can be found in mortality tables. The company collects a premium of \)250 each year as payment for the insurance. The amount Y that the company earns on a randomly selected policy of this type is \(250 per year, less the \)100,000 that it must pay if the insured dies. Here is the probability distribution of Y:

(a) Explain why the company suffers a loss of $98,750 on such a policy if a client dies at age 25.

(b) Calculate the expected value of Y. Explain what this result means for the insurance company.

(c) Calculate the standard deviation of Y. Explain what this result means for the insurance company.

The given information is:

If the customer died at the age of 25, the company had to pay out the face value of $100,000 because the client had paid the premium five times in total (as the client had paid the premium yearly since he was 21).

The company's profit was then calculated as the total of the 5 premiums (250 each premium) minus the cost of 100,000. Profit:

$$\begin{gathered} =5x\$250-\$100,000 \\ =\$1,250-\$100,000 \\ =-\$98,750 \end{gathered}$$

$$\begin{gathered} \mu =\sum xP\left(X=x\right) \\ =\left(-99,750\right)\times \left(0.00183\right)+\left(-99,500\right)\times \left(0.00186\right)+\left(-99,250\right)\times \left(0.00189\right)+ \\ \left(-99,000\right)\times \left(0.00191\right)+\left(98,750\right)\times \left(0.00193\right)+\left(1250\right)\times \left(0.995058\right) \\ =303.3525 \end{gathered}$$

The average profit on the randomly selected policy of the firm is $303.3525

The given mean is 303.3525.

The predicted value of the squared departure from the mean is known as the variance:

$$\begin{gathered} {\sigma }^2=\sum {\left(x-\mu \right)}^{2}P\left(x\right) \\ ={\left(-99750-303.3525\right)}^2\times \left(0.00183\right)+{\left(-99500-303.3525\right)}^2\times \left(0.00186\right)+ \\ {\left(99250-303.3525\right)}^2\times \left(0.00189\right)+{\left(99000-303.3525\right)}^2\times \left(0.00191\right)+ \\ {\left(98750-303.3525\right)}^2\times \left(0.00193\right)+{\left(1250-303.3525\right)}^2\times \left(0.99058\right) \\ \approx 94,236,827 \end{gathered}$$

The standard deviation is:

$$\begin{gathered} \sigma =\sqrt{{\sigma }^2} \\ =\sqrt{94,236,827} \\ \approx 9707.57 \end{gathered}$$