The average turkey sold in the U.S. in 2005 weighed about 28 pounds, with a standard deviation of 3 pounds. The average in 1965 was 18 pounds, with a standard deviation of 2 pounds. At which time would it be more unusual to have a 23 pound turkey?

2005: \(\dfrac{23-28}3\approx-1.67\)

1965: \(\dfrac{23-18}2=2.5\)

Since the 23 lb turkey is closer to the 2005 mean and further from the 1965 mean, it would be more unusual to find such a turkey in 1965. (The closer the above values (z-scores) are to zero, the more likely they are to occur.)

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slicergiza slicergiza · Answer 2 · 2019-09-02T04:58:24-04:00

Answer:

In 1965

Step-by-step explanation:

If in the normal distribution, z-scores is less than -1.96 or higher than 1.96 then it is considered to be unusual.

The mean weight in 2005, \(\mu_1=28\)

Standard deviation, \(\sigma_1=3\)

Also, if x represents average turkey sold ( in pounds ),

Then x = 23

So, z-score would be,

\(z_1 =\frac{x-\mu_1}{\sigma_1}\)

\(=\frac{23-28}{3}\)

\(=-\frac{5}{3}\)

\(\approx -1.67\)

Now, mean weight in 165, \(\mu_2=18\)

\(\sigma=2\)

So, the z-score would be,

\(z_2=\frac{23-18}{2}=\frac{5}{2}=2.5\)

\(\because | -1.96 - z_1 | < |1.96 - z_2 |\)

Hence, in 1965 it will be more unusual to have a 23 pound turkey.