Question

Suppose babies born in a large hospital have a mean weight of 3215 grams, and a variance of 84,681. if 67 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by less than 52 grams? round your answer to four decimal places.

Answer 1

The z-score is given by:
z=(x-μ)/σ
but
x-μ=52
thus
z=52/√84681
z=0.1787
Thus:
P(X≤52)=0.9633

Answer 2

The correct answer is:

0.8558.

Explanation:

The weights could be higher or lower than the mean by 52 grams.  

Our formula for a z-score is:
$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}$

We have the variance instead of the standard deviation.  Variance is the square of the standard deviation; to find the standard deviation, we take the square root:
√84681 = 291 = σ

We already have X-μ; this is 52 or -52.  This gives us two z-scores:
$z=\frac{-52}{\frac{291}{\sqrt{67}}}\text{ or } z=\frac{52}{\frac{291}{\sqrt{67}}} \\ \\z=-1.46\text{ or }z=1.46$

Using a z-table, the area under the curve to the left of (less than) -1.46 is 0.0721. 
The area under the curve to the left of (less than) 1.46 is 0.9279.

We want the area between these two scores, since the difference between the weight and the mean could be less than or equal to 52; this means we subtract these probabilities:
0.9279-0.0721 = 0.8558.