Answer:
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
<u>So, the standard error of x’ using the finite population correction factor is give by</u>
σ(x’) = √[(N - n)/(N - 1)] x (sigma/√n)
σ(x’) = √[(4000 - 40)/(4000 - 1)] x (8.2/√40)
σ(x’) = 1.29
<u>Standard error of x’ without using the finite population correction factor is</u>
σ(x’) = σ/√n
σ(x’) = 8.2/√40 = 1.2965
<u>There is little difference between the two values of the standard error. So we can ignore the population correction factor.</u>
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
<u>So the required probability is </u>
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
