04/08/23 Section 12.4 1

Section 12.4The Normal Distribution

Objectives

1. Recognize characteristics of normal distributions.

2. Understand the 68-95-99.7 Rule.

3. Find scores at a specified number of standard deviations from the mean.

4. Use the 68-95-99.7 Rule

5. Convert a data item to a z-score.

6. Understand percentiles and quartiles.

7. Solve applied problems involving normal distributions.

8. Use and interpret margins of error.

9. Recognize distributions that are not normal.

04/08/23 Section 12.4 2

Normal Distribution

• Also called the bell curve or Gaussian distribution.• Normal distribution is bell shaped and symmetric about a

vertical line through its center.• Mean, median and mode are all equal and located at the

center of the distribution.

04/08/23 Section 12.4 3

Normal Distribution continued

The shape of the normal distribution depends on the mean and the standard deviation. These three graphs have the same mean but different standard deviations. As the standard deviation increases, the distribution becomes more spread out.

04/08/23 Section 12.4 4

Standard Deviation and the 68-95-99.7 Rule1. Approximately 68% of

the data items fall within 1 standard deviation of the mean (in both directions).

2. Approximately 95% of the data items fall within 2 standard deviations of the mean.

3. Approximately 99.7% of the data items fall within 3 standard deviations of the mean.

04/08/23 Section 12.4 5

Example 1Finding Scores at a Specified Standard

Deviation From the Mean

• Male adult heights in North America are approximately normally distributed with a mean of 70 inches and a standard deviation of 4 inches. Find the height that is 2 standard deviations above the mean.

• Solution. • Height = mean + 2∙standard deviation

= 70 + 2∙4 = 70 + 8 = 78

04/08/23 Section 12.4 6

Example 2Using the 68-95-99.7 Rule

Mean – 1∙standard deviation

= 70 - 1∙ 4= 66.

Mean + 1∙standard deviation

= 70 + 1∙ 4 = 74.

68% of males have heights between 66 and 74 inches.

• Use the distribution of male adult heights in the figure to find the percentageof men in North America withheights between 66 inchesand 74 inches.• Solution. The 68-95-99.7 Rule states That approximately 68% of the data items fall within 1 standard deviation, 4, of the mean, 70.

04/08/23 Section 12.4 7

Computing z-Scores in Normal Distributions.

• Describes how many standard deviations a data item in a normal distribution lies above or below the mean. Computing z-scores:

z-score

• Data items above the mean have positive z-scores. • Data items below the mean have negative z-scores.• The z-score for the mean is 0.

deviation standardmean - item data

04/08/23 Section 12.4 8

Example 3 Computing z-Scores

• The mean weight of newborn infants is 7 pounds and the standard deviation is 0.8 pound.The weights of newborn infantsare normally distributed. Find thez-score for a weight of 9 pounds.

• Solution:

The mean is 7 and the

standard deviation is

0.8. The z-score

written z9, is:

2.5 0.8

2

0.8

79

deviation standard

mean - item data9

z

04/08/23 Section 12.4 9

Example 4Understanding Scores

• Intelligence quotients (IQs) on the Stanford – Binet

intelligence test are normally distributed with a mean of 100

and a standard deviation of 16. What is the IQ

corresponding to a z-score of -1.5?• Solution:

The negative sign in -1.5 tells us that the IQ is 1½ standard

deviations below the mean.

Score = mean – 1.5 ∙ standard deviation

= 100 – 1.5 (16) = 100 – 24 = 76.

The IQ corresponding to a z-score of -1.5 is 76.

04/08/23 Section 12.4 10

Percentiles and Quartiles

• Percentiles: If n% of the itemsin a distribution are less than aparticular data item, we say that the data item is in the nth percentile of the distribution.• Quartiles: Divide data sets into four equal parts. The 25th percentile is the first quartile.25% of the data fall below the firstquartile. The 50th percentile is the second quartile. The 75th percentile is the third quartile.

04/08/23 Section 12.4 11

A Percentile Interpretation for z-Scores

• Using the z-score and a table, you can find the percentage of data items that are less than any data item in a normal distribution.

• In a normal distribution, the mean, median, and mode all have a corresponding z-score of 0 and are the 50th percentile. Thus, 50% of the data items are greater than or equal to the mean, median and mode.

04/08/23 Section 12.4 12

Example 5Finding the Percentage of Data Items Less Than a

Given Data Item

• According to the Department of Health and Education,

cholesterol levels are normally distributed. For men

between 18 and 24 years, the mean is 178.2 and the

standard deviation is 40.7. What percentage of men in this

age range have a cholesterol level less than 239.15?

• Solution

Compute the z-score for a 239.15 cholesterol level.

1.540.7

61.05

40.7

178.1239.15

deviation standard

mean - item data239.15

z

04/08/23 Section 12.4 13

Example 5 continued

• We must find the percentage of men with a cholesterol level less than z = 1.5. The table gives this percentage asa percentile. Finding 1.5 in the z-score column gives a percentile of 93.32. thus , 93.32% of men between 18 and 24 have a cholesterol level less than 239.15

z-score Percentile

1.4 91.92

1.5 93.32

1.6 94.52

04/08/23 Section 12.4 14

Finding the Percentage of Data Items Between Two Given Data Items.

1. Convert each given data item to a z-score.

2. Use the table to find the percentile corresponding to each z-score in step 1.

3. Subtract the lesser percentile from the greater percentile and attach a % sign.

04/08/23 Section 12.4 15

Example 6• The amount of time that self-employed Americans workeach week is normally distributed with a mean of 44.6 hoursand a standard deviation of 14.4 hours. What percentage ofself-employed individuals in the United States work between 37.4 and 80.6 hours per week?Solution:Step 1. Convert each given data item to a z-score.

0.514.4

7.2-

14.4

44.6 37.4

deviation standard

mean - item data37.4

z

2.514.436

14.444.680.6

deviation standardmean - item data

80.6

z

04/08/23 Section 12.4 16

Example 6 continued

Step 2. Use the Table to find the percentile correspondingto these z-scores.

The percentile corresponding to -0.50 is 30.85.

This means that 30.85 percent of self-employed Americans work fewer than 37.4 hours per week.

The Table also shows that the percentile that corresponds

to a z-score of 2.5 is 99.38.

That means that 99.38% of self- employed Americans work fewer than 80.6 hours per week.

04/08/23 Section 12.4 17

Example 6 continued

Step 3. Subtract the lesser percentile from the greater

percentile and attach a % sign.

99.38 – 30.85 = 68.53.

Thus, 68.53% of self-employed

Americans work between

37.4 and 80.6 hours per week.

= 68.53 %

04/08/23 Section 12.4 18

Summary of Computing Percentage of Data Items for Normal Distributions

Description of Percentage Graph Computation of PercentagePercentage of data Use the table percentile

items less than a given for z = b and add a % sign

data item with z = b

Percentage of data Subtract the table percentile

items greater than a for z = a from 100 and addgiven data item with a % sign.

z = a

Percentage of data Subtract the table percentile

items between two for z = a from the table

given data items with percentile for z = b and add

z = a and z = b a % sign.

04/08/23 Section 12.4 19

Polls and Margins of Error

• Statisticians use properties of the normal distribution to estimate the probability that a result obtained from a single sample reflects what is truly happening.

• If n is the sample size, there is a 95% probability that it lies within of the true population statistic.

• is called the margin of error.

n

1

n

1

04/08/23 Section 12.4 20

Example 7Using and Interpreting Margin of Error

In a random sample of 1172 children ages 6 through 14, 17% of the children said getting bossed around is a bad thing about being a kid.a. Verify the margin of error.Solution: The sample size is n = 1172. The margin of error is

b. Write a statement about the percentage of children who feel that getting bossed around is a bad thing about being a kid.

Solution: There is a 95% probability that the true population percentage lies between

17% -2.9% = 14.1% and 17% + 2.9% = 19.9%

2.9%1172

1

n

1

04/08/23 Section 12.4 21

Other Kinds of Distributions

• This graph represents

the population distribution

of weekly earnings in the

United States. There is no

upper limit on weekly earnings.

The relatively few people with very high weekly incomes pull

the mean income to a value greater than the median. • The most frequent income, the mode, occurs towards the

low end of the data items. • This is called a skewed distribution because a large number

of data items are piled up at one end or the other with a “tail” at the other end.

• This graph is skewed to the right.