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An insurance company claims that in the entire population of homeowners, the mean annual loss from fire is --$250 and the standard deviation of the loss is ơ-$1000. The distribution of losses is strongly right-skewed: many policies have $0 loss, but a few have large losses. An auditor examines a random sample of 10,000 of the company's policies. If the company's clairm is correct, what's the probability that the average loss from fire in the sample is no greater than $275?

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Answer:

the probability that the average loss from fire in the sample is no greater than $275 is 0.9938

Step-by-step explanation:

given information:

mean, μ = $250

std deviation, σ = $1000

random sample, n = 10000

x = $275

P([tex]x[/tex] ∠_ 275) = P(z < (z ∠ (x - μ)/(σ/√n))

                    = P (z ∠ (275 - 250)/(1000/√10000)

                    = P(z ∠ 2.5)

                     =  0.9938

The probability that the average loss from fire in the sample is no greater than $275 is 99.38%.

Given to us,

  • mean, μ = $250
  • standard deviation, σ = $1000
  • random sample, n = 10000
  • Average loss, x ≤ $275

To find

the probability that the average loss from fire in the sample is no greater than $275,

[tex]P(z\leq x)=P[z< \dfrac{(x-\mu )}{(\dfrac{\sigma}{\sqrt n})}][/tex]

[tex]P(z\leq 275)=P[z< \dfrac{(275-250 )}{(\dfrac{1000}{ \sqrt {10000}})}][/tex]

[tex]P(z\leq 275)=P[z< \dfrac{(25 )}{(10)}][/tex]

[tex]P(z\leq 275)=P[z< {(2.5)}][/tex]

[tex]P(z\leq 275)=0.9938[/tex]

P(x ≤ $275) = 99.38%

Hence, the probability that the average loss from fire in the sample is not greater than $275 is 99.38%.

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