The owner of a meat market has an assistant who has determined that the weights of roasts are normally distributed, with a mean of 3.2 pounds and standard deviation of 0.8 pounds. If a sample of 25 roasts yields a mean of 3.6 pounds, what is the Z-score for this sample mean?
Group of answer choices
1. None of these choices.
2. −2.50
3. . 2.50
4. −0.50
Population mean,
= 3.2 pounds
Population standard deviation,
= 0.8 pounds
Sample size, n = 25
For sampling distribution of mean,
Mean,
=
= 3.2 pounds
Standard error,
=
=
= 0.16
Here,
= 3.6 pounds
Z score of sample mean = (
-
)/
= (3.6 - 3.2)/0.16
= 2.5
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