The Normal Distribution

Chapter 2

Continuous Random Variable

• A continuous random variable:– Represented by a function/graph.– Area under the curve represents the proportions of

the observations– Total area is exactly 1.

• How do we locate the median for a continuous random variable? the mean?

• The median is the value that divides the graph into equal area while the mean is the “balance” point.

Continuous Random Variable 1

What percent of the observations lie below 0.4?

0.50.4

A= .4(1) =0.4

40%

Continuous Random Variable 2

What proportion of the observations lie above 0.6?

0.6

A= 1.4(.5) =0.7

Notice, to find proportion for observation above, we can use the complement rule.

Continuous Random Variable 3

• Where is the mean and median?• How will the curve change as changes?

Normal Distributions

• Symmetric, single-peaked, and mound-shaped distributions are called normal distributions

• Normal curves:– Mean = median– The mean and standard deviation completely

determine the shape

The Normal Curve

• Will finding proportions work different than previous random variable examples?

• Empirical Rule Discovery

68% of observations fall within 1 of

95% of observations fall within 2 of

99.7% of observations fall within 3 of

68-95-99.7 Rule

Applet

68-95-99.7 Rule

34% 34%

13.5%13.5%2.35% 2.35%.15%

Applet

Percentiles?

34% 34%

13.5%13.5%2.35% 2.35%

50th 84th 16th

What’s Normal in Statistics?

• Normal distributions are good descriptions for real data allowing measures of relative position to be easily calculated (i.e. percentiles)

• Much of statistical inference (in this course) procedures area based on normal distributions

• FYI: many distributions aren’t normal

Distribution of dates is approximately normal with mean 1243 and standard deviation of 36 years.

1243 1279 1315 135112071135 1171

Assume the heights of college women are normally distributed with a mean of 65 inches and standard deviation of 2.5 inches.

65 67.5 70 72.562.557.5 60

What percentage of women are taller than 65 in.?

65 67.5 70 72.562.557.5 60

50%

What percentage of women are shorter than 65 in.?

65 67.5 70 72.562.557.5 60

50%

What percentage of women are between 62.5 in. and 67.5 in.?

65 67.5 70 72.562.557.5 60

68%

What percentage of women are between 60 in. and 70 in.?

65 67.5 70 72.562.557.5 60

95%

What percentage of women are between 60 and 67.5 in?

65 67.5 70 72.562.557.5 60

68%

13.5%

81.5%

What percentage of women are shorter than 70 in.?

65 67.5 70 72.562.557.5 60

50%

34%

97.5% 13.5%

Iliana’s Grade

• After 5 weeks of class Iliana must transfer from a stat class at Lanier to this class. Last week was the chapter 1 test in both classes. Iliana scored a 61 out of 70. Let’s say our test was out of 100 points. What score should she be given?

Iliana’s claim

• Iliana claims that her test at Lanier was harder than our test.

• Does your previous method of assigning a grade take in consideration difficulty?

• If we have all of the data, what important facts can we utilize to improve our assignment of Iliana’s grade?

Important Facts

• Maximum possible on our test was 100 pts while Lanier’s test was 70 pts.

• Mean score on Lanier’s test was 50.5 pts while our test was 77.2 pts.

• Standard deviation on Lanier’s test was 5.3 pts while ours was 8.1 pts.

• Test scores from both high schools tend to be normally distributed.

• How will we fairly assign Iliana’s score?

Lanier’s distribution

50.5 55.8 61.1 66.4

Iliana

Lanier’s and McCallum’s distributions

50.5 55.8 61.1 66.4

Iliana

77.2 85.3 93.4 101.5

Iliana

•How can we find Iliana’s relative position?

Iliana’s score – class average standard deviation

Formula

• What is the formula to find the relative position for any distribution?

Iliana’s score – class average standard deviation

z–score=x

1. Suppose as student has taken two quizzes in a statistics course. On the first quiz the mean score was 32, the standard deviation was 8, and the student received a 44. The student obtained a 28 on the second quiz, for which the mean was 23 and the standard deviation was 3. If test scores are approximately normal, on which quiz did the student perform better relative to the rest of the class?

zx

First quiz: z 44 328 15.

Second quiz: z 28 233

1667.

Relative to their departments, which is better paid if the husband earns $14.60 and the wife earns $15.75?

z

14 60 12 80

12015

. ..

.Husband:

Wife: z

15 75 1350

180125

. ..

.

3. A married couple is employed by the same company. The husband works in a department for which the mean hourly rate is $12.80 and the standard deviation is $1.20. His wife is employed in a department where the mean rate is $13.50 and the standard deviation is $1.80.

z 15.93nd percentile

What percentile is the husband located in his department?

z 125. P z( . ) . . 125 1 8944 1056

What percent of employees in the wife’s department earn better than her?

What would the wife need to earn to match her husband’s relative position?

z

14 60 12 80

12015

. ..

.Husband:

Wife: z

15 75 1350

180125

. ..

.

x

1350180

15.

.. x 1350 2 7. .

x $16.20

The wife would need to earn $16.20 to match the husband’s relative position.

If the husband wanted to earn in the 95th percentile, how much should he earn per hour?

Need a z-score of 1.65!

z

14 60 12 80

12015

. ..

.

x 12 80120

..

165.

x 12 80 198. .

x $14.78

The husband will need to earn at least $14.78 to be in the 95th percentile.

2

– 2

14.24 .0228P x 59.60 .9772P x

14.24 59.60 .9772 .0228 .9544P x

89%

44.5%

5.5%

z-score = –1.60 z-score = 1.60

94.5%

36.921.60

11.32

x

36.921.60

11.32

x

The middle 89% of the data ranges from 18.81 to 55.03 ppb.

Warm Up Answer Check

• Probability of a president being 52 year old or younger is .3192.

• Probability of SAT math score less than 540 is .8907.

• Probability of Aprils having one or fewer frost days is .2389.

• Probability of hotdogs having over 500 mg of sodium is .2177.

.3488

.8627

.4

.2593

PREZ

Data is approximately normal.

SATM

Data is fairly normal.

FROST

Data is not normal.

HOTDG

Data is approximately normal.