# Men have head breadths that are normally distributed with a mean of 6.0 inches and a standard deviation of 1.0 inches. If one male is randomly selected, what is the probability that his head breadth is less than 6.2 inches?

## The Safeguard Helmet Company plans an initial production run of 1200 helmets. What is the probability that 100 randomly selected men have a mean head breadth less than 6.2 inches?

Nov 23, 2017

.5793 or 57.93%
.9772 or 97.72%

#### Explanation:

We want to find the probability that, in a normally distributed series, we will encounter a value that is less than the value that is .2 greater than the mean.

The z-table is great for this kind of problem.

The z-score for a head breadth of 6.2 would be .2 because the mean is 6, and the standard deviation is 1.

$\frac{\mathrm{de} v i a t i o n}{\sigma} = z$

$\frac{.2}{1} = .2$ (If the standard deviation were a number other than 1, we would need to do this to find the z-score of a deviation from the mean)

We can now look in our positive z-table (the one I've linked here is from chegg.com) to find the area of a normal curve to the left of $z = 0.20$, and we see that it is .5793, meaning that there is a 57.93% chance that you will stumble upon a man with a head breadth under 6.2 inches. Edit: I see that this is a two-part question.

For the second part, you want to use this formula:

${\sigma}_{\overline{x}} = \frac{\sigma}{\sqrt{n}}$

with $n$ being the sample size.

$\frac{\sigma}{\sqrt{n}} \to \frac{1.0}{\sqrt{100}}$

${\sigma}_{\overline{100}} = 0.1$

We can now use the central limit theorem.

$z = \frac{\overline{x} - {\mu}_{\overline{x}}}{\sigma} _ \overline{x}$

$z = \frac{6.2 - 6.0}{0.1}$

$z = 2.00$

Going back to our z table:

$z = 2.00 \to .9772$

or a 97.72% chance.