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.

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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