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3 years ago

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Seventy million pounds of trout are grown in the U.S. every year. Farm-raised trout contain an average of 32 grams of fat per po

und, with a standard deviation of 7 grams of fat per pound. A random sample of 34 farm-raised trout is selected. The mean fat content for the sample is 31.1 grams per pound. Find the probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout.

1 answer:

LenaWriter [7]3 years ago

7 0

Answer:

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout is 0.227

Step-by-step explanation:

Let X be the sample mean of fat in 34 farm-raised trout

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout can be stated as:

P(X<31.1 grams) = P(z<z*)

where z* is the z-statistic of sample mean of 31.1 grams fat.

z* can be calculated as follows:

z*= $\frac{X-M}{\frac{s}{\sqrt{N} } }$ where

X = 31.1 grams

M is the average grams of fat per pound (32)

s is the standard deviation (7)

N is the sample size (34)

Then z*= $\frac{31.1-32}{\frac{7}{\sqrt{34} } }$ ≈ −0.7497

and P(z<z*) ≈ 0.227

The probability of observing a sample mean of 31.1 grams of fat per pound or less in a random sample of 34 farm-raised trout is 0.227

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Answer:

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Step-by-step explanation:

To solve for the value of each loan we will use the formula:

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Now let's move on to the Security bank.

The security bank charges 7.5% exact interest. For exact interest we use 365 days.

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