An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is and the standard deviation of the loss is $\mathrm{\sigma }=\mathrm{\backslash (}5000.$The distribution of losses is strongly right-skewed: many policies have $\backslash )0$loss, but a few have large losses. The company hopes to sell $1000$ of these policies for $\backslash (300$each.

a. Assuming that the company’s claim is true, what is the probability that the mean loss from fire is greater than $\mathrm{\backslash )}300$for an SRS of $1000$ homeowners?

b. 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?

The mean is $\mathrm{\mu }=250$and standard deviation is $\mathrm{\sigma }=5000$

The number of homeowners is $\mathrm{n}=1000$

The sample mean $\overline{\mathrm{x}}=300$

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

The sampling distribution of the sample mean $\overline{\mathrm{x}}$is also normal because the population distribution is normal.

The Z-score is

Using the normal probability, the associating probability is calculated.

$\mathrm{P}(\mathrm{Z}<0.32)$s typical normal probability table in the appendix has a row beginning with $0.3$ and a column beginning with $0.2.$

$\begin{array}{rcl}P(\overline{X}& \ge & 300)=P(Z>0.32)\\ & =& 1-P(Z<0.32)\\ & =& 1-0.6255\\ & =& 0.3745\\ & =& 37.45\%\end{array}$

The mean is $\mathrm{\mu }=250$and standard deviation is $\mathrm{\sigma }=5000$

The number of homeowners is $\mathrm{n}=1000$

$\mathrm{P}(\overline{\mathrm{X}}\le \overline{\mathrm{x}})=90\mathrm{\%}$

The following concept was used:

$\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}$

Determine the z-score in the normal probability table that corresponds to a probability of $90$percent or $0.90$, and the closest probability is width="51">0.8997, which is located in the normal probability table's row $1.2$ and column$.08$, and hence the equivalent z-score is $1.2+.08=1.28$.

localid="1657629267213" $\mathrm{z}=1.28$

Z-score is localid="1657629270630" $\mathrm{z}=\frac{\mathrm{x}-{\mathrm{\mu }}_{\overline{\mathrm{x}}}}{{\mathrm{\sigma }}_{\overline{\mathrm{x}}}}=\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}=\frac{\overline{\mathrm{x}}-250}{5000\sqrt{1000}}$

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

localid="1657629274503" $\begin{array}{r}\frac{\overline{\mathrm{x}}-250}{5000/\sqrt{1000}}=1.28\\ \overline{\mathrm{x}}-250=1.28(5000/\sqrt{1000})\end{array}$

To each side, add $250$

localid="1657629377013" $\overline{\mathrm{x}}-250+1.28(5000/\sqrt{1000})$

Determine:

localid="1657629382095" $\mathrm{x}=452.39$

As a result, the business should charge is localid="1657629389088" $\mathrm{\$}452.39.$