2. to estimate the mean age for a population of 4000 employees, a simple random sample of 40 employees is selected. a. would you use the finite population correction factor in calculating the standard error of the mean? explain. b. if the population standard deviation is σ = 8.2 years, compute the standard error both with and without the finite population correction factor. c. what is the probability that the sample mean age of the employees will be within ±2 years of the population mean age?

Step-by-step explanation:

a.

Size of the population, N = 4000

Size of the sample, n = 40

n/N = 40/4000 = 0.01

0.01 is less than 0.05 and hence we would not us the finite population correction factor in calculating the standard error of the mean.

b.

Population standard deviation is σ = 8.2

So, the standard error of x’ using the finite population correction factor is give by

σ(x’) = √[(N - n)/(N - 1)] x (sigma/√n)

σ(x’) = √[(4000 - 40)/(4000 - 1)] x (8.2/√40)

σ(x’) = 1.29

Standard error of x’ without using the finite population correction factor is

σ(x’) = σ/√n

σ(x’) = 8.2/√40 = 1.2965

There is little difference between the two values of the standard error. So we can ignore the population correction factor.

c.

Let the population mean be μ

Probability that the sample mean will be within =-2 of the population mean is

P(μ– 2 < x’ < μ + 2)

At x’ = μ – 2 , we have

z = (μ – 2 – μ)/1.2965

z = -1.54

at x’ = μ  + 2, we have

z = (μ + 2 – μ)/1.2965

z = 1.54

So the required probability is

P(μ – 2 < x’ < μ + 2) = p(-1.54< z < 1.54)

P(μ – 2 < x’ < μ + 2) = p(z < 1.54) – p(z < -1.54)

P(μ – 2 < x’ < μ + 2) = 0.9382 – 0.0618

P(μ – 2 < x’ < μ + 2) = 0.8764

the solution is 12x-6y