Answer to Question #163811 in Statistics and Probability for Zafar

Question

An insurance company knows that the average insured value of the house policies it holds is $253255 with a standard deviation of $40648. It takes a random sample of 289 policies to investigate if insurance is adequate in light of recent natural disasters and finds that the probability is 0.15 that the average insured value of this sample exceeds an amount A.

Using the Excel formula, calculate the value of A correct to the nearest cent.

Answer 1

$P(Z>z)=0.15.$

$z=$ 1.04.

$\frac{A-\mu}{\frac{\sigma}{\sqrt{n}}}=z.$

$A=\mu+z\frac{\sigma}{\sqrt{n}}=253255+1.04\frac{40648}{\sqrt{289}}=255741.70.$