### Math 225 Course Notes

### Section 4.7: Normal Distribution Applications

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.

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.

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:

- of more than 200
- less than 100
- 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.

Last modified: Feb 7, 1996

Bret Larget,
larget@mathcs.duq.edu