Question

In the EAI sampling problem, the population mean is $51,000 and the population standard deviation is $4,000 . When the sample size is n=20, there is a .4972 probability of obtaining a sample mean +- 600 within of the population mean. Use z-table.

a. What is the probability that the sample mean is within $600 of the population mean if a sample of size 40 is used (to 4 decimals)?

b. What is the probability that the sample mean is within of the population mean if a sample of size 80 is used (to 4 decimals)?

Answer 1

a)
Here, μ = 51000, σ = 4000/sqrt(40) = 632.4555, x1 = 50400 and x2 = 51600. We need to compute P(50400<= X <= 51600). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (50400 - 51000)/632.4555 = -0.95
z2 = (51600 - 51000)/632.4555 = 0.95

Therefore, we get
P(50400 <= X <= 51600) = P((51600 - 51000)/632.4555) <= z <= (51600 - 51000)/632.4555)
= P(-0.95 <= z <= 0.95) = P(z <= 0.95) - P(z <= -0.95)
= 0.8289 - 0.1711
= 0.6578

b)
Here, μ = 51000, σ = 4000/sqrt(80) = 447.2136,

z = (x - μ)/σ
z1 = (50400 - 51000)/447.2136 = -1.34
z2 = (51600 - 51000)/447.2136 = 1.34

Therefore, we get
P(50400 <= X <= 51600) = P((51600 - 51000)/447.2136) <= z <= (51600 - 51000)/447.2136)
= P(-1.34 <= z <= 1.34) = P(z <= 1.34) - P(z <= -1.34)
= 0.9099 - 0.0901
= 0.8198