Question

The owner of a fish market has an assistant who has determined that the weights of catfish are normally distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. What percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds

Answer 1

Answer:

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

Step-by-step explanation:

For a normal random variable with mean Mu = 3.2 and standard deviation sd = 0.8 there is a distribution of the sample mean (MX) for samples of size 4, given by:

Z = (MX - Mu) / sqrt (sd ^ 2 / n) = (MX - 3.2) / sqrt (0.64 / 4) = (MX - 3.2) / 0.4

For a sample mean of 3.0, Z = (3 - 3.2) / 0.4 = -0.5

For a sample mean of 3.0, Z = (4 - 3.2) / 0.4 = 2.0

P (3.2 <MX <4) = P (-0.5 < Z <2.0) = 0.6687.

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is 66.87%

Answer 2

The percentage of samples of 4 fish will have sample means between 3.0 and 4.0 pounds is; 66.78%

What is sampling distribution?

We are given;

mean; μ = 3.2

Standard deviation; σ = 0.8

sample size; n = 4

z-score is;

z = (x'- μ)/(√σ²/n)

z = (x' - 3.2)/(√0.8²/4)  

z = (x' - 3.2)/0.4

For a sample mean of 3.0,

z = (3 - 3.2)/0.4

z = -0.5

For a sample mean of 4.0,

Z = (4 - 3.2)/0.4

z = 2.0

P (3.2 < x' <4) = P (-0.5 < Z <2.0) = 0.6687 or 66.87%

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