Question

1. The mean income of households in the US is $37922 with standard deviation $12500. Suppose we take a sample of 400 households. Find the probability that the sample mean of these households is: a) Less than $36800 b) (*) Between $36800 and $39700 c) Within $1100 of the population mean

Answer 1

a)

Here, μ = 37922, σ = 625 and x = 36800. We need to compute P(X <= 36800). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (36800 - 37922)/625 = -1.8

Therefore,
P(X <= 36800) = P(z <= (36800 - 37922)/625)
= P(z <= -1.8)
= 0.0359

b)

Here, μ = 37922, σ = 625, x1 = 36800 and x2 = 39700. We need to compute P(36800<= X <= 39700). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (36800 - 37922)/625 = -1.8
z2 = (39700 - 37922)/625 = 2.84

Therefore, we get
P(36800 <= X <= 39700) = P((39700 - 37922)/625) <= z <= (39700 - 37922)/625)
= P(-1.8 <= z <= 2.84) = P(z <= 2.84) - P(z <= -1.8)
= 0.9977 - 0.0359
= 0.9618

c)

Here, μ = 37922, σ = 625, x1 = 36822 and x2 = 39022. We need to compute P(36822<= X <= 39022). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (36822 - 37922)/625 = -1.76
z2 = (39022 - 37922)/625 = 1.76

Therefore, we get
P(36822 <= X <= 39022) = P((39022 - 37922)/625) <= z <= (39022 - 37922)/625)
= P(-1.76 <= z <= 1.76) = P(z <= 1.76) - P(z <= -1.76)
= 0.9608 - 0.0392
= 0.9216