Despite its nutritional value, seafood is only a tiny pan of the American diet,

Question

Despite its nutritional value, seafood is only a tiny pan of the American diet, with the average American eating just 16 pounds of seafood per year. Ms. Lauren and Ms. Samantha both work in the seafood industry and they decide to create their own random samples and document the average seafood diet in their sample. Let the standard deviation of the American seafood diet be 7 pounds. a. Ms. Lauren samples 42 Americans and finds an average seafood consumption of 18 pounds. How likely is it to get an average of 18 pounds or more if she had a representative sample? b. Ms. Samantha samples 90 Americans and finds an average seafood consumption of 17.5 pounds. How likely is it to get an average of 17.5 pounds or more if she had a representative sample? c. Which of the two women is likely to have used a more representative sample? Explain.

Explanation / Answer

Z-score Problems

Solution:

We are given the population average = 16

Population standard deviation = 7

a)      Here, we are given the sample size n = 42

Sample mean = 18

Here, we have to find the P(X>18)

P(X>18) = 1 – P(X<18)

Now, we have to find P(X<18) by using the z-score formula

The z-score formula is given as below:

Z = (sample mean – population mean) / [population standard deviation / sqrt (n)]

Now, plug all the values in the above formula and find the z-score given as below:

Z = (18 – 16) / [7/sqrt(42)]

Z = 1.85164

P(Z<18) = P(Z<1.85164) = 0.967961

P(X>18) = 1 – P(X<18) = 1 – P(Z<1.85164) = 1 – 0.967961 = 0.032039

It is 3.2% likely to get an average of 18 pounds or more if she had a representative sample.

b)      Here, we are given a sample size n = 90 and sample mean = 17.5

We have to find P(X>17.5)

P(X>17.5) = 1 – P(X<17.5)

We have to find P(X<17.5) by using the z score formula.

The z-score formula is given as below:

Z = (sample mean – population mean) / [population standard deviation / sqrt (n)]

Now, plug all the values in the above formula and find the z-score given as below:

Z = (17.5 – 16) / [7/sqrt(90)]

Z = 2.032893

P(X>17.5) = 1 – P(X<17.5) = 1 – P(Z<2.032893) = 1 – 0.978968 = 0.021032

It is 2.1% likely to get an average of 17.5 pounds or more if she had a representative sample.

c)       Here, we have to following information:

It is 3.2% likely to get an average of 18 pounds or more if Ms. Lauren had a representative sample.

It is 2.1% likely to get an average of 17.5 pounds or more if Ms. Samantha had a representative sample.

So, it is observed that Ms. Lauren used more representative sample.

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