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 a 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 ample mean?

Answer 1

Answer:

2.5

Step-by-step explanation:

We are given that

Mean=$\mu=3.2$ pounds

Standard deviation=$\sigma=0.8$ pounds

n=25

We have to find the Z-score if the mean of  a sample of 25 roasts is 3.6 pounds.

We know that Z-score formula

$Z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n}}}$

We have $\bar X=3.6$

Substitute the values then we get

Z- score=$\frac{3.6-3.2}{\frac{0.8}{\sqrt{25}}}$

Z- score=$\frac{0.4}{\frac{0.8}{5}}=\frac{0.4\times 5}{0.8}=2.5$

Hence, the Z-score value for the given sample mean=2.5