# Answer to Question #12597 in Statistics and Probability for Kolton

Question #12597

Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of grams of fat per pound, with a standard deviation of grams of fat per pound. A random sample of farm-raised trout is selected. The mean fat content for the sample is grams per pound. Find the probability of observing a sample mean of grams of fat per pound or less in a random sample of farm-raised trout.

Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.

Expert's answer

For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)

You can translate into standard normal units by:

Z = ( X - μ ) / σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed

with mean μ and standard deviation σ /√(n)

In this question we have

Xbar ~ Normal( μ = 32 , σ² = 56.25 / 36 )

Xbar ~ Normal( μ = 32 , σ² = 1.5625 )

Xbar ~ Normal( μ = 32 , σ = 7.5 / sqrt( 36 ) )

Xbar ~ Normal( μ = 32 , σ = 1.25 )

Find P( Xbar < 29.7 )

P( ( Xbar - μ ) / σ < ( 29.7 - 32 ) / 1.25 ) = P( Z < -1.84 ) = 0.03288412.

You can translate into standard normal units by:

Z = ( X - μ ) / σ

Where Z ~ Normal( μ = 0, σ = 1). You can then use the standard normal cdf tables to get probabilities.

If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.

If a sample of size is is drawn from a population with mean μ and standard deviation σ then the sample average xBar is normally distributed

with mean μ and standard deviation σ /√(n)

In this question we have

Xbar ~ Normal( μ = 32 , σ² = 56.25 / 36 )

Xbar ~ Normal( μ = 32 , σ² = 1.5625 )

Xbar ~ Normal( μ = 32 , σ = 7.5 / sqrt( 36 ) )

Xbar ~ Normal( μ = 32 , σ = 1.25 )

Find P( Xbar < 29.7 )

P( ( Xbar - μ ) / σ < ( 29.7 - 32 ) / 1.25 ) = P( Z < -1.84 ) = 0.03288412.

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