+Modeling Distributions of Data

2.1 Describing Location in a Distribution

2.2 Normal Distributions

+

Normal Distributions

After this section, you should be able to…

DESCRIBE and APPLY the 68-95-99.7 Rule

DESCRIBE the standard Normal Distribution

PERFORM Normal distribution calculations

ASSESS Normality

Learning Objectives

+Norm

al Distributions

Normal Distributions One particularly important class of density curves are the

Normal curves, which describe Normal distributions. All Normal curves are symmetric, single-peaked, and bell-

shaped

A Specific Normal curve is described by giving its mean µ and standard deviation σ.

Two Normal curves, showing the mean µ and standard deviation σ.

+Density curvesA density curve is a mathematical model of a distribution.

The total area under the curve, by definition, is equal to 1, or 100%.

The area under the curve for a range of values is the proportion of all observations for that range.

Histogram of a sample with the smoothed, density curve

describing theoretically the population.

+

Density curves come in any imaginable

shape.

Some are well known mathematically and

others aren’t.

+

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

A family of density curves

Here means are different

( = 10, 15, and 20) while

standard deviations are the same

( = 3)

Here means are the same ( = 15)

while standard deviations are

different ( = 2, 4, and 6).

+Describing Location in a D

istribution Density Curves

In Chapter 1, we developed a kit of graphical and numerical tools for describing distributions. Now, we’ll add one more step to the strategy.

1. Always plot your data: make a graph.2. Look for the overall pattern (shape, center, and spread) and

for striking departures such as outliers.3. Calculate a numerical summary to briefly describe center

and spread.

Exploring Quantitative Data

4. Sometimes the overall pattern of a large number of observations is so regular that we can describe it by a smooth curve.

+Describing Location in a D

istribution Density Curve

Definition:

A density curve is a curve that•is always on or above the horizontal axis, and•has area exactly 1 underneath it.

A density curve describes the overall pattern of a distribution. The area under the curve and above any interval of values on the horizontal axis is the proportion of all observations that fall in that interval.

The overall pattern of this histogram of the scores of all 947 seventh-grade students in Gary, Indiana, on the vocabulary part of the Iowa Test of Basic Skills (ITBS) can be described by a smooth curve drawn through the tops of the bars.

+ Normal DistributionsN

ormal D

istributions

Definition:

A Normal distribution is described by a Normal density curve. Any particular Normal distribution is completely specified by two numbers: its mean µ and standard deviation σ.

•The mean of a Normal distribution is the center of the symmetric Normal curve.

•The standard deviation is the distance from the center to the change-of-curvature points on either side.

•We abbreviate the Normal distribution with mean µ and standard deviation σ as N(µ,σ).

Normal distributions are good descriptions for some distributions of real data.

Normal distributions are good approximations of the results of many kinds of chance outcomes.

Many statistical inference procedures are based on Normal distributions.

+Norm

al Distributions

Although there are many Normal curves, they all have properties in common.

The 68-95-99.7 Rule

Definition: The 68-95-99.7 Rule (“The Empirical Rule”)

In the Normal distribution with mean µ and standard deviation σ:

•Approximately 68% of the observations fall within σ of µ.

•Approximately 95% of the observations fall within 2σ of µ.

•Approximately 99.7% of the observations fall within 3σ of µ.

99.7%

+Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)

For data with a (symmetric) bell-shaped distribution, the standard deviation has the following characteristics:

11

• About 68% of the data lie within one standard deviation of the mean.

• About 95% of the data lie within two standard deviations of the mean.

• About 99.7% of the data lie within three standard deviations of the mean.

+Interpreting Standard Deviation: Empirical Rule (68 – 95 – 99.7 Rule)

12

3x s x s 2x s 3x sx s x2x s

68% within 1 standard deviation

34% 34%

99.7% within 3 standard deviations

2.35% 2.35%

95% within 2 standard deviations

13.5% 13.5%

+

mean µ = 64.5 standard deviation = 2.5

N(µ, ) = N(64.5, 2.5)

All Normal curves N) share the same properties

Reminder: µ (mu) is the mean of the idealized curve, while is the mean of a sample.

s (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.

About 68% of all observations

are within 1 standard deviation

(of the mean ().

About 95% of all observations

are within 2 of the mean .

Almost all (99.7%) observations

are within 3 of the mean.

Inflection point

x

+

Because all Normal distributions share the same properties, we can

standardize our data to transform any Normal curve N() into the

standard Normal curve N(0,1).

The standard Normal distribution

For each x we calculate a new value, z (called a z-score).

N(0,1)

=>

z

x

N(64.5, 2.5)

Standardized height (no units)

+

N(0,1)

z (x )

The cool thing about working with normally distributed data is that we can manipulate it and then find answers to questions that involve comparing seemingly non-comparable distributions.

We do this by “standardizing” the data. All this involves is changing the scale so that the mean now = 0 and the standard deviation =1. If you do this to different distributions it makes them comparable.

+Norm

al Distributions

The distribution of Iowa Test of Basic Skills (ITBS) vocabulary scores for 7th grade students in Gary, Indiana, is close to Normal. Suppose the distribution is N(6.84, 1.55).

a) Sketch the Normal density curve for this distribution.

b) What percent of ITBS vocabulary scores are less than 3.74?

c) What percent of the scores are between 5.29 and 9.94?

+Norm

al Distributions

The Standard Normal Distribution All Normal distributions are the same if we measure in units

of size σ from the mean µ as center.

Definition:

The standard Normal distribution is the Normal distribution with mean 0 and standard deviation 1.If a variable x has any Normal distribution N(µ,σ) with mean µ and standard deviation σ, then the standardized variable

has the standard Normal distribution, N(0,1).

z x -

+Describing Location in a D

istribution Measuring Position: z-Scores

A z-score tells us how many standard deviations from the mean an observation falls, and in what direction.

Definition:

If x is an observation from a distribution that has known mean and standard deviation, the standardized value of x is:

A standardized value is often called a z-score.

zx mean

standard deviation

Jenny earned a score of 86 on her test. The class mean is 80 and the standard deviation is 6.07. What is her standardized score?

zx mean

standard deviation

86 80

6.070.99

+Describing Location in a D

istribution Using z-scores for Comparison

We can use z-scores to compare the position of individuals in different distributions.

Jenny earned a score of 86 on her statistics test. The class mean was 80 and the standard deviation was 6.07. She earned a score of 82 on her chemistry test. The chemistry scores had a fairly symmetric distribution with a mean 76 and standard deviation of 4. On which test did Jenny perform better relative to the rest of her class?

zstats 86 80

6.07

zstats 0.99

zchem 82 76

4

zchem 1.5

+Using Table A

(…)

Table A gives the area under the standard Normal curve to the left of any z value.

.0082 is the area under N(0,1) left

of z = -2.40

.0080 is the area under N(0,1) left

of z = -2.41

0.0069 is the area under

N(0,1) left of z = -2.46

+Norm

al Distributions

The Standard Normal Table

Because all Normal distributions are the same when we standardize, we can find areas under any Normal curve from a single table.

Definition: The Standard Normal Table

Table A is a table of areas under the standard Normal curve. The table entry for each value z is the area under the curve to the left of z.

Z .00 .01 .02

0.7 .7580 .7611 .7642

0.8 .7881 .7910 .7939

0.9 .8159 .8186 .8212

P(z < 0.81) = .7910

Suppose we want to find the proportion of observations from the standard Normal distribution that are less than 0.81. We can use Table A:

+Norm

al Distributions

Finding Areas Under the Standard Normal Curve

Find the proportion of observations from the standard Normal distribution that are between -1.25 and 0.81.

Can you find the same proportion using a different approach?

1 - (0.1056+0.2090) = 1 – 0.3146

= 0.6854

+Norm

al Distributions

Normal Distribution Calculations

State: Express the problem in terms of the observed variable x.

Plan: Draw a picture of the distribution and shade the area of interest under the curve.

Do: Perform calculations.

•Standardize x to restate the problem in terms of a standard Normal variable z.•Use Table A and the fact that the total area under the curve is 1 to find the required area under the standard Normal curve.

Conclude: Write your conclusion in the context of the problem.

How to Solve Problems Involving Normal Distributions

mean µ = 64.5"

standard deviation = 2.5" x (height) = 67"

We calculate z, the standardized value of x:

mean from dev. stand. 1 15.2

5.2

5.2

)5.6467( ,

)(

zx

z

Because of the 68-95-99.7 rule, we can conclude that the percent of women

shorter than 67” should be, approximately, .68 + half of (1 - .68) = .84 or 84%.

Area= ???

Area = ???

N(µ, ) = N(64.5, 2.5)

= 64.5” x = 67”

z = 0 z = 1

Ex. Women heights

Women heights follow the N(64.5”,2.5”)

distribution. What percent of women are

shorter than 67 inches tall (that’s 5’6”)?

Area ≈ 0.84

Area ≈ 0.16

N(µ, ) =

N(64.5”, 2.5”)

= 64.5” x = 67” z = 1

Conclusion:

84.13% of women are shorter than 67”.

By subtraction, 1 - 0.8413, or 15.87% of

women are taller than 67".

For z = 1.00, the area under

the standard Normal curve

to the left of z is 0.8413.

Percent of women shorter than 67”

+Tips on using Table A

Because the Normal distribution is

symmetrical, there are 2 ways that

you can calculate the area under the

standard Normal curve to the right of

a z value.

area right of z = 1 - area left of z

Area = 0.9901

Area = 0.0099

z = -2.33

area right of z = area left of -z

+Tips on using Table ATo calculate the area between 2 z- values, first get the area under N(0,1)

to the left for each z-value from Table A.

area between z1 and z2 =

area left of z1 – area left of z2

A common mistake made by

students is to subtract both z

values. But the Normal curve

is not uniform.

Then subtract the

smaller area from the

larger area.

The area under N(0,1) for a single value of z is zero

(Try calculating the area to the left of z minus that same area!)

The National Collegiate Athletic Association (NCAA) requires Division I athletes to

score at least 820 on the combined math and verbal SAT exam to compete in their

first college year. The SAT scores of 2003 were approximately normal with mean

1026 and standard deviation 209.

What proportion of all students would be NCAA qualifiers (SAT ≥ 820)?

16%. approx.or

0.1611 is .99- z

ofleft the toN(0,1)

under area :A Table

99.0209

206209

)1026820(

)(

209

1026

820

z

z

xz

x

Note: The actual data may contain students who scored exactly 820 on the SAT. However, the proportion of scores exactly equal to 820 is 0 for a normal distribution is a consequence of the idealized smoothing of density curves.

area right of 820 = total area - area left of 820= 1 - 0.1611

≈ 84%

The NCAA defines a “partial qualifier” eligible to practice and receive an athletic

scholarship, but not to compete, as a combined SAT score is at least 720.

What proportion of all students who take the SAT would be partial

qualifiers? That is, what proportion have scores between 720 and 820?

7%. approx.or

0.0721 is .99- z

ofleft the toN(0,1)

under area :A Table

46.1209

306209

)1026720(

)(

209

1026

720

z

z

xz

x

About 9% of all students who take the SAT have scores

between 720 and 820.

area between = area left of 820 - area left of 720 720 and 820 = 0.1611 - 0.0721

≈ 9%

What is the effects of better maternal care on gestation time and premies?

The goal is to obtain pregnancies 240 days (8 months) or longer.

Ex. Gestation time in malnourished mothers

What improvement did we get

by adding better food?

0.3085. is 0.5- z ofleft the

toN(0,1)under area :A Table

deviation) standard a (half

5.020

1020

)250240(

)(

20

250

240

z

z

xz

x

Vitamins Only

Under each treatment, what percent of mothers failed to carry their babies at

least 240 days?

Vitamins only: 30.85% of women

would be expected to have gestation

times shorter than 240 days.

=250, =20, x=240

0.0418. is 1.73- z ofleft the

toN(0,1)under area :A Table

mean) from sd 2almost (

73.115

2615

)266240(

)(

15

266

240

z

z

xz

x

Vitamins and better food

Vitamins and better food: 4.18% of women

would be expected to have gestation times

shorter than 240 days.

=266, =15, x=240

Compared to vitamin supplements alone, vitamins and better food resulted in a much

smaller percentage of women with pregnancy terms below 8 months (4% vs. 31%).

+Norm

al Distributions

Normal Distribution CalculationsWhen Tiger Woods hits his driver, the distance the ball travels can be described by N(304, 8). What percent of Tiger’s drives travel between 305 and 325 yards?

When x = 305, z =305 - 304

80.13

When x = 325, z =325 - 304

82.63

Using Table A, we can find the area to the left of z=2.63 and the area to the left of z=0.13.0.9957 – 0.5517 = 0.4440. About 44% of Tiger’s drives travel between 305 and 325 yards.

+Norm

al Distributions

Assessing Normality The Normal distributions provide good models for some

distributions of real data. Many statistical inference procedures are based on the assumption that the population is approximately Normally distributed. Consequently, we need a strategy for assessing Normality.

Plot the data.

•Make a dotplot, stemplot, or histogram and see if the graph is approximately symmetric and bell-shaped.

Check whether the data follow the 68-95-99.7 rule.

•Count how many observations fall within one, two, and three standard deviations of the mean and check to see if these percents are close to the 68%, 95%, and 99.7% targets for a Normal distribution.

+Norm

al Distributions

Normal Probability Plots Most software packages can construct Normal probability plots.

These plots are constructed by plotting each observation in a data set against its corresponding percentile’s z-score.

If the points on a Normal probability plot lie close to a straight line, the plot indicates that the data are Normal. Systematic deviations from a straight line indicate a non-Normal distribution. Outliers appear as points that are far away from the overall pattern of the plot.

Interpreting Normal Probability Plots

+

Describing Location in a Distribution

After this section, you should be able to…

INTERPRET cumulative relative frequency graphs

MEASURE position using z-scores

TRANSFORM data

DEFINE and DESCRIBE density curves

Learning Objectives

+

Example: Using the Empirical Rule

In a survey conducted by the National Center for Health Statistics, the sample mean height of women in the United States (ages 20-29) was 64 inches, with a sample standard deviation of 2.71 inches. Estimate the percent of the women whose heights are between 64 inches and 69.42 inches.

37

+

Solution: Using the Empirical Rule

38

3x s x s 2x s 3x sx s x2x s55.87 58.58 61.29 64 66.71 69.42 72.13

34%

13.5%

• Because the distribution is bell-shaped, you can use the Empirical Rule.

34% + 13.5% = 47.5% of women are between 64 and 69.42 inches tall.

+Example: Comparing z-Scores from Different Data Sets

In 2007, Forest Whitaker won the Best Actor Oscar at age 45 for his role in the movie The Last King of Scotland. Helen Mirren won the Best Actress Oscar at age 61 for her role in The Queen. The mean age of all best actor winners is 43.7, with a standard deviation of 8.8. The mean age of all best actress winners is 36, with a standard deviation of 11.5. Find the z-score that corresponds to the age for each actor or actress. Then compare your results.

39

+Solution: Comparing z-Scores from Different Data Sets

40

Forest Whitaker

45 43.70.15

8.8

xz

• Helen Mirren61 36

2.1711.5

xz

0.15 standard deviations above the mean

2.17 standard deviations above the mean

+Solution: Comparing z-Scores from Different Data Sets

41

The z-score corresponding to the age of Helen Mirren is more than two standard deviations from the mean, so it is considered unusual. Compared to other Best Actress winners, she is relatively older, whereas the age of Forest Whitaker is only slightly higher than the average age of other Best Actor winners.

z = 0.15 z = 2.17

+Describing Location in a D

istribution Describing Density Curves

Our measures of center and spread apply to density curves as well as to actual sets of observations.

The median of a density curve is the equal-areas point, the point that divides the area under the curve in half.

The mean of a density curve is the balance point, at which the curve would balance if made of solid material.

The median and the mean are the same for a symmetric density curve. They both lie at the center of the curve. The mean of a skewed curve is pulled away from the median in the direction of the long tail.

Distinguishing the Median and Mean of a Density Curve

+Normal Distributions

In this section, we learned that…

The Normal Distributions are described by a special family of bell-shaped, symmetric density curves called Normal curves. The mean µ and standard deviation σ completely specify a Normal distribution N(µ,σ). The mean is the center of the curve, and σ is the distance from µ to the change-of-curvature points on either side.

All Normal distributions obey the 68-95-99.7 Rule, which describes what percent of observations lie within one, two, and three standard deviations of the mean.

Summary

+Normal Distributions

In this section, we learned that…

All Normal distributions are the same when measurements are standardized. The standard Normal distribution has mean µ=0 and standard deviation σ=1.

Table A gives percentiles for the standard Normal curve. By standardizing, we can use Table A to determine the percentile for a given z-score or the z-score corresponding to a given percentile in any Normal distribution.

To assess Normality for a given set of data, we first observe its shape. We then check how well the data fits the 68-95-99.7 rule. We can also construct and interpret a Normal probability plot.

+Inverse normal calculationsWe may also want to find the observed range of values that

correspond to a given proportion/ area under the curve.

For that, we use Table A backward: For an area to the left of 1.25 % (0.0125),

we first find the desired

area/ proportion in the

body of the table,

we then read the

corresponding z-value

from the left column and

top row.

For an area to the left of 1.25 % (0.0125), the z-value is -2.24

25695.255

)15*67.0(266

)*()(

0.67.-about is N(0,1)

under 25% arealower

for the valuez :A Table

?

%25arealower

%75areaupper

15

266

x

x

zxx

z

x

Vitamins and better food

=266, =15, upper area 75%

How long are the longest 75% of pregnancies when malnutritioned mothers are

given vitamins and better food?

?

upper 75%

The 75% longest pregnancies in this group are about 256 days or longer.

Remember that Table A gives the area to

the left of z. Thus we need to search for

the lower 25% in Table A in order to get z.

Summary

the z-value is -2.24