Answer:
Percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.20 kg = 78.5%
The sampling error is 0.08083, in terms of the sampling error, 78.5% of samples of three men will have mean brain weights within (1.24×sampling error) of the mean.
Explanation:
Complete Question
According to one study, brain weights of men are normally distributed with mean = 1.20 kg and a standard deviation = 0.14 kg.
Determine the percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.20 kg. Interpret your answer in terms of sampling error.
Solution
The Central limit theorem allows us to say
The mean of sampling distribution is approximately equal to the population mean.
μₓ = μ = 1.20 kg
And the standard deviation of the sampling distribution is given as
σₓ = (σ/√N)
σ = population standard deviation = 0.14 kg
N = sample size = 3
σₓ = (0.14/√3) = 0.08083
Percentage of all samples of three men that have mean brain weights within 0.1 kg of the population mean brain weight of 1.20 kg, that is, percentage of all samples of three men with mean brain weights within 1.10 kg and 1.30 kg.
P(1.10 ≤ x ≤ 1.30)
We first normalize or standardize 1.10 and 1.30
The standardized score for any value is the value minus the mean then divided by the standard deviation.
For 1.10 kg
z = (x - μₓ)/σₓ = (1.10 - 1.20)/0.08083 = -1.24
For 1.30 kg
z = (x - μₓ)/σₓ = (1.30 - 1.20)/0.08083 = 1.24
To determine the required probability
P(1.10 ≤ x ≤ 1.30) = P(-1.24 ≤ z ≤ 1.24)
We'll use data from the normal distribution table for these probabilities
P(1.10 ≤ x ≤ 1.30) = P(-1.24 ≤ z ≤ 1.24)
= P(z ≤ 1.24) - P(z ≤ -1.24)
= 0.89251 - 0.10749
= 0.78502 = 78.502%
The sampling error is 0.08083, in terms of the sampling error, 78.5% of samples of three men will have mean brain weights within (1.24×sampling error) of the mean.
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