Question

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

Answer 1

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