Answer:
a) P(21.18 ≤ x ≤ 22.18) = 0.87644
b) P(18.30 ≤ x ≤ 19.30) =0.91456
c) P(x > 18.50) = 0.9452
Step-by-step explanation:
The sample mean is related to the population mean through
μₓ = μ
For the male graduates,
μₓ = μ = $21.68
For female graduates,
μₓ = μ = $18.80
Standard deviation for the male graduates = $2.30
Standard deviation for the female graduates = $2.05
And the standard deviation of the distribution of sample means is related to the population standard deviation through
σₓ = σ/√n
where n = Sample size
a) probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?
Standard deviation of the distribution of sample means = σₓ = (2.30/√50) = $0.325
sample means that are $0.5 of the population mean range from ($21.18 and $22.18) (that is, mean ± $0.5)
Required probability = P(21.18 ≤ x ≤ 22.18)
This is a normal distribution problem
We first normalize $21.18 and $22.18
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For $21.18,
z = (x - μ)/σ = (21.18 - 21.68)/0.325 = -1.54
For $22.18,
z = (x - μ)/σ = (22.18 - 21.68)/0.325 = 1.54
Required probability
P(21.18 ≤ x ≤ 22.18) = P(-1.54 ≤ z ≤ 1.54)
We'll use data from the normal probability table for these probabilities
P(21.18 ≤ x ≤ 22.18) = P(-1.54 ≤ z ≤ 1.54)
= P(z ≤ 1.54) - (P ≤ -1.54)
= 0.93822 - 0.06178
= 0.87644
b) What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?
Standard deviation of the distribution of sample means = σₓ = (2.05/√50) = $0.290
sample means that are $0.5 of the population mean range from ($18.30 and $19.30)
Required probability = P(18.30 ≤ x ≤ 19.30)
This is a normal distribution problem
We first normalize $18.30 and $19.30
For $18.30
z = (x - μ)/σ = (18.30 - 18.80)/0.290 = -1.72
For $19.30,
z = (x - μ)/σ = (19.30 - 18.80)/0.290 = 1.72
Required probability
P(18.30 ≤ x ≤ 19.30) = P(-1.72 ≤ z ≤ 1.72)
We'll use data from the normal probability table for these probabilities
P(18.30 ≤ x ≤ 19.30) = P(-1.72 ≤ z ≤ 1.72)
= P(z ≤ 1.72) - (P ≤ -1.72)
= 0.95728 - 0.04272
= 0.91456
What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?
σₓ = (2.05/√120) = $0.187
sample means that are more $0.3 below the population mean range from ($18.50)
Required probability = P(x > 18.50)
This is a normal distribution problem
We first normalize $18.50
z = (x - μ)/σ = (18.50 - 18.80)/0.187 = -1.60
Required probability
P(x > 18.50) = P(z > -1.60)
We'll use data from the normal probability table for these probabilities
P(x > 18.50) = P(z > -1.60) = 1 - P(z ≤ -1.60)
= 1 - 0.05480 = 0.9452
Hope this Helps!!!
