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