Problem #3 (12 points) infor insurance per household in the $40,000. mation from

Question

Problem #3 (12 points) infor insurance per household in the $40,000. mation from the American Institute of Insurance United States has a mean $110,000 and standard deviation indicates that the amount of life For all possible random samples of 100 households, what is the shape of the distribution of the sample means? a. sampling A point) b. Specify the mean and standard devíation of the sampling distribution of the sampke values on the template below. means, for samples of size 100 and mark appropriate 3 points) c. What is the likelilhood that in a random sample of 100 households the mean cost of wte (4 points) insurance is more than $100,000? at is the probability that in a random sample of 100 households the mean cost of We 4 points) insurance is below $120,000? Show your work. Partial credit will be given

Explanation / Answer

Solution

Let X = Amount (in $)of life insurance per household in the US.

We are given Mean of X = µ = 110000 ………………………….…… (A) and

Standard Deviation of X = ? = 40000 ………………………….…… (B)

Back-up Theory

If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then,

Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X ? or ? t) = P[{(X - µ)/?} ? or ? {(t - µ)/?}] = P[Z ? or ? {(t - µ)/?}] .………(2)

Part (a)

The sampling distribution of the sample mean based on a sample of size 100 would be bell shaped, by virtue of Central Limit Theorem. ANSWER 1

Part (b)

If the sample mean of 100 values = Xbar, then

Mean (Xbar) = Mean (X) = 110000 ANSWER 2 [vide (A)]

Standard Deviation, SD(Xbar) = {SD(X)}/?n = 40000/10 = 4000 ANSWER 3

[vide (B) and n = 100]

Part (c)

Likely-hood of the mean cost being more than 100000

= P(Xbar > 100000)

= P[Z > {(100000- 1100000/4000}] [vide (2)]

= P(Z > - 2.5)

= 0.9938 ANSWER 4

Part (d)

Probability of the mean cost being below 120000

= P(Xbar < 120000)

= P[Z < {(120000- 1100000/4000}] [vide (2)]

= P(Z < 2.5)

= 0.9938 ANSWER 5

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