Question

Assume that men's weights are normally distributed with a mean of 172 lb and standard deviation of 29 lb (national health survey) if 81 men are randomly selected, find the probability that they have a mean weight less than 167 lb.

Answer 1

Answer: 0.0606

Step-by-step explanation:

Given : The men's weights are normally distributed with a mean of 172 lb and standard deviation of 29 lb.

i.e. $\mu=172$ and $\sigma=29$

and sample size : n= 81

Let x be a random variable that denotes the men's weights.

Formula : $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Then, the probability that they have a mean weight less than 167 lb will be :-

$P(x$

Hence, the  probability that they have a mean weight less than 167 lb = 0.0606

Answer 2

The probability is 0.0606.

We find the z-score associated with this sample using the formula:

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

Using our information, we have:
$z=\frac{167-172}{\frac{29}{\sqrt{81}}}=-1.55$

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this score is 0.0606.