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Question

An insurance company claims that in the entire population of homeowners the mean annual loss from fire is μ=$250\mu=\$ 250 and the standard deviation of the loss is σ=$5000.\sigma=\$ 5000. The distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. The company hopes to sell 1000 of these policies for$300 each. If the company wants to be 90% certain that the mean loss from fire in an SRS of 1000 homeowners is less than the amount it charges for the policy, how much should the company charge?

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Given:

μ=Mean=250σ=Standard deviation=5000n=Sample size=1000P(Xx)=90%\begin{align*} \mu&=\text{Mean}=250 \\ \sigma&=\text{Standard deviation}=5000 \\ n&=\text{Sample size}=1000 \\ P(\overline{X}\leq \overline{x})&=90\% \end{align*}

Let us determine the z-score that corresponds with a probability of 90% or 0.90 in the normal probability table of the appendix. We note that the closest probability is 0.8997 which lies in the row 1.21.2 and in the column .08 of the normal probability table and thus the corresponding z-score is then 1.2+.08=1.281.2+.08=1.28.

z=1.28z=1.28

The sampling distribution of the sample mean x\overline{x} has mean μ\mu and standard deviation σn\dfrac{\sigma}{\sqrt{n}}.

The z-score is the value decreased by the mean, divided by the standard deviation.

z=xμxσx=xμσ/n=x2505000/1000z=\dfrac{\overline{x}-\mu_{\overline{x}}}{\sigma_{\overline{x}}}=\dfrac{\overline{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{\overline{x}-250}{5000/\sqrt{1000}}

The two found expressions of the z-score then have to be equal:

x2505000/1000=1.28\dfrac{\overline{x}-250}{5000/\sqrt{1000}}=1.28

Multiply each side by 5000/10005000/\sqrt{1000}:

x250=1.28(5000/1000)\overline{x}-250=1.28(5000/\sqrt{1000})

Add 250 to each side:

x=250+1.28(5000/1000)x=250+1.28(5000/\sqrt{1000})

Evaluate:

x=452.39x= 452.39

Thus the company should charge $452.39.

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