### Section 4.7: Normal Distribution Applications

#### Key Concepts

In many (but not all) instances, the distribution of a variable is well approximated by a normal curve. This section shows how to use the standard normal table from the previous section to solve problems where the mean and standard deviation are not standard. The variable must be standardized.

#### Standardization

The z-score tells the number of standard deviations an observation x is from the mean, where a negative z-score implies that x is below the mean. z-scores are found by
```  z = (x-mu)/sigma
```
If the z-score is known and the value of x is needed, solving the previous equation for x gives
```  x = mu + z * sigma
```
This simply states that x is z standard deviations above the mean.

#### Examples

Solve Exercise 4.7.2.

Ridge counts in finger prints are approximately normally distributed with a mean mu = 140 and standard deviation sigma = 50.

Find the probability that an individual chosen randomly has a ridge count:

1. of more than 200
2. less than 100
3. between 100 and 200
Solutions:

Always draw a sketch.

(1) The area to the right of 200 under the given normal curve is equal to the area to the right of

```  z = (200 - 140)/50 = 1.20
```
under the standard normal curve. The area to the left of 1.20 is .8849 from the table, so the answer is 1 - .8849 = .1151.

(2) The area to the left of 100 under the given normal curve is equal to the area to the left of

```  z = (100 - 140) / 50 = -0.80
```
under the standard normal curve. This area is .2119 from the table.

(3) The area between 100 and 200 under the given normal curve is equal to the area between -0.80 and 1.20 under the standard normal curve. From the above solutions, this is .8849 - .2119 = .6730.