Question

An insurance company sells homeowner's insurance to compensate homeowners in case of damage to their home. Suppose we assume that the probability distribution of x = insurance company's payout on all policies is severely right-skewed with a mean of $4,750 and a standard deviation of $18,839.7850.

1. Consider a random sample of 9 homeowners who own this policy. Find the mean and standard deviation of the average insurance payout on these 9 policies.

µ _x̄ = $____ (no commas)

σ _x̄ = $____     (4 decimal places, no commas)

2. Consider a random sample of 2,500 homeowners who own this policy. Find the mean and standard deviation of the average insurance payout on these 2,500 policies.

µ _x̄ = $_____ (no commas)

σ _x̄ = $_____ (4 decimal places)

3. Which sampling scenario would permit using the normal curve as a model for the sampling distribution of x̄?

A.only the sample of size 9
B.only the sample of size 2,500
C.both sample sizes
D.neither sample sizes

4. Which of these best states the Central Limit Theorem?

A.For random samples, as the sample size increases, the shape of the sample data approaches a normal distribution.
B.For random samples, as the sample size increases, the shape of the distribution of x̄ approaches a normal distribution.
C.For any sampling method, as the sample size increases, the shape of the distribution of x̄ approaches a normal distribution.
D.For random samples, as the population size increases, the shape of the sample data approaches a normal distribution.
E.For any sampling method, as the sample size increases, the value of x̄ approaches µ.
F.For random samples, as the sample size increases, the shape of the population data approaches a normal distribution. 5. For the sampling scenario with 2,500 homeowners, find the probability that the average payout exceeds $6,000. _____(5 decimal places) 6. Suppose you randomly selected 2,500 homeowners with this policy and found that the average payout was $7,000. What would you conclude?

Answer 1

1)

µ x̄ = 4750

σ x̄ =18839.7850/sqrT(9)=6279.9283

2)

µ x̄ =4750

σ x̄ =18839.7850/sqrT(2500)=6279.9283=376.7957

3)

B.only the sample of size 2,500

4)

B.For random samples, as the sample size increases, the shape of the distribution of x̄ approaches a normal distribution.

5)

probability that the average payout exceeds $6,000:

probability =

P(X>6000)

=

P(Z>3.32)=

1-P(Z<3.32)=

1-0.9995=

0.0005

6)

as probabiltiy of that event is almost negligible ; therefore we may assume that there is a chance of having a higher mean value of  payout