Section 2.2 ~ Standard Normal Calculations
Recall from section 2.1:

In section 2.1 density curves were introduced:

A density curve is an idealized mathematical model that describes the overall pattern of a distribution


The area under the curve represents a proportion of all observations and is therefore equal to 1
More specifically, we looked at the basic layout of the normal curve

Symmetric, bell shaped, unimodal distribution in which the mean is equal to the median

The 68-95-99.7 rule was used to help describe observations within 3 standard deviations above and
below the mean

The notation used for a normal distribution is in the form N(mean, standard deviation) or N (  ,  )
The standard normal distribution:

Has a mean of _______ and a standard deviation of ________.
 Using the correct notation: N (0,1)
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Standardizing is when you: Take any normal distribution and convert it to have a mean of 0 and StDev of 1

A standardized values is called: a z-score

A z-score tells us: how many standard deviations an observation falls away from the mean, and in which
direction.


Observations larger than the mean: have positive z-scores

Observations smaller than the mean have negative z-scores.
To standardize a value: subtract the mean of the distribution from the observation and then divide by the
standard deviation.
z
x

Example 1 – Standardizing a Value
The heights of young women are approximately normal with   64.5 inches and   2.5 inches.
a. What is the standardized height for a woman that is 68 inches tall?
z
68  64.5
 1.4
2.5
This means that a woman who is 68 inches tall has a standardized height that is 1.4 deviations above the mean
b. What is the standardized height for a woman that is 5 feet tall?
z
60  64.5
 1.8
2.5
A woman who is 5 feet tall has a standardized height that is 1.8 deviations below the mean
Normal Distribution Calculations:

Recall that the area under a density curve represents a proportion of observations.

Since all normal distributions are the same when standardized:

We can use a single table to find any area under the curve


Table A – first page of textbook.
The standard normal table will always give: the area to the left of the z-score
Example 2 – Using the Standard Normal Table
Using the z-scores found in example 1, determine the proportion of women who are:
a. Shorter than 68 inches:
z
68  64.5
 1.4
2.5
P( x  68)  .9192
The proportion of young women that are shorter than 68 inches is 91.92%
b. Taller than 5 ft.:
Since the table gives the area below the z-score, that area needs to be subtracted from 1 in order to find the
area above.
z
60  64.5
 1.8
2.5
P( x  60)  .0359
P( x  60)  1  .0359  .9641
The proportion of young women that are taller than 5 ft. is 96.41%.
Steps to follow for a normal distribution problem:
1. State the problem in terms of the observed variable x. Draw a picture of the distribution and shade the area of
interest under the curve.
2. Standardize x to restate the problem in terms of a z-score. On the picture label the Z-score.
3. Find the required area under the standard normal curve by using table A, and the fact that the total area under
the curve is 1.
4. Write your conclusion in the context of the problem.
Example 3 – Is Cholesterol a Problem With Young Boys?
The level of cholesterol in the blood is important because high cholesterol levels may increase risk of heart disease. The
distribution of blood cholesterol levels in a large population of people of the same age and sex are roughly normal. For
14-year-old boys, the mean is   170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation
is   30 mg/dl. 3 Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more
than 240 mg/dl of cholesterol?

State the problem:
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Call the level of cholesterol in the blood x.
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The variable x has the N(170, 30) distribution.
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We want the proportion of boys with cholesterol level x > 240 – P ( x  240)
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Standardize x and draw a picture:
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Use the table:
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From the table we see that the proportion of observations less than 2.33 (or 240mg/dl) is .9901

Thus, to the right of 2.33 is 1 - .9901 = .0099


This is about .01 or 1%
Write your conclusion in the context of the problem:

Only about 1% of boys have high cholesterol
Example 4:
What percent of 14-year-old boys have blood pressure between 164 and 220 mg/dl?