Question

1. The file name Final.xlsx, sheet named Part 4is kept blank. Use Part 4sheet to answer this question. Put a box around your answers. Round your answer to 4 decimal places.

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $110,000 with a standard deviation of $40,000. Assume the population distribution is normal. A random sample of 100 households is taken.  

1. What is the probability that sample mean will be less than $120,000?
2. What is the probability that sample mean will be between $105,000 and $120,000?

PLEASE HELP ME WITH A AND B .

Answer 1

Given,

Population mean = μ = 110000

Population standard deviation = σ = 40000

Sample size = 100

Hence, sample mean = μ_s = μ = 110000

Sample standard deviation = σ_s = σ/√n = 40000/√100 = 4000

(a) Let the probability that the sample mean is less than x = 120000 be P(z)

for normal distribution, z = (x - μ_s) / σ_s = (120000 - 110000)/4000 = 2.5

from z table, for z = 2.5, P(z) = 0.9938

Hence, probability that the sample mean will be less than 120000 = 99.38%

(b)

Let the probability that the sample mean is less than x₁ = 105000 be P(z₁)

for normal distribution, z₁ = (x₁ - μ_s) / σ_s = (105000 - 110000)/4000 = -1.25

from z table, for z₁ = 1.25, P(z₁) = 0.1056

from part (a), Probability that the sample mean is less than x₂ = 120000 = P(z₂) = 0.9938

Hence, Probability that the sample mean is between 105000 and 120000 = P(z₂) - P(z₁) = 0.9938 - 0.1056 = 0.8882 or 88.82%