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 $x_2=\mu +1.96\sigma$ .

Figure 14

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

$50\pm 1.96\times 2=50\pm 3.92\text{mm}$

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:

Answer

Now deduce the corresponding values $x_1,x_2$ for the general normal distribution:

Answer

Next, find the values for $x_1$ and $x_2$ for the given problem:

Answer

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

Answer