### 8 Probability intervals - general normal distribution

We saw in subsection 3 that $95\%$ of the area under the standard normal curve lay between ${z}_{1}=-1.96$ and ${z}_{2}=1.96$ . Using the formula $Z=\frac{X-\mu}{\sigma}$ in the re-arrangement $X=\mu +Z\sigma $ . We can see that $95\%$ of the area under the general normal curve lies between ${x}_{1}=\mu -1.96\sigma $ and ${x}_{2}=\mu +1.96\sigma $ .

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Figure 14
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##### Example 14

Suppose that the internal diameters of mass-produced pipes are normally distributed with mean 50 mm and standard deviation 2 mm. What are the $95\%$ probability limits on the internal diameter of a single pipe?

##### Solution

Here $\mu =50$ $\sigma =2$ so that the $95\%$ probability limits are

$\phantom{\rule{2em}{0ex}}50\pm 1.96\times 2=50\pm 3.92\text{mm}$

i.e. 46.08 mm and 53.92 mm.

The probability interval is (46.08, 53.92).

##### Task!

What is the $99\%$ probability interval for the lifetime of a bulb when the lifetimes of such bulbs are normally distributed with a mean of 2000 hours and standard deviation of 40 hours?

First sketch the standard normal curve marking the values ${z}_{1},{z}_{2}$ between which $99\%$ of the area under the curve is located:

Now deduce the corresponding values ${x}_{1},{x}_{2}$ for the general normal distribution:

${x}_{1}=\mu -2.58\sigma ,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{x}_{2}=\mu +2.58\sigma $

Next, find the values for ${x}_{1}$ and ${x}_{2}$ for the given problem:

$\phantom{\rule{2em}{0ex}}{x}_{1}=2000-2.58\times 40=1896.8\phantom{\rule{1em}{0ex}}\text{hours}$

$\phantom{\rule{2em}{0ex}}{x}_{2}=2000+2.58\times 40=2103.2\phantom{\rule{1em}{0ex}}\text{hours}$

Finally, write down the $99\%$ probability interval for the lifetimes:

(1896.8 hours, 2103.2 hours).