+Chapter 2Modeling Distributions of Data

2.1 Describing Location in a Distribution

2.2 Normal Distributions

+Section 2.1Describing Location in a Distribution

After this section, you should be able to…

MEASURE position using percentiles

INTERPRET cumulative relative frequency graphs

MEASURE position using z-scores

TRANSFORM data

DEFINE and DESCRIBE density curves

Learning Objectives

+ Measuring Position: Percentiles One way to describe the location of a value in a distribution

is to tell what percent of observations are less than it.

Definition:

The pth percentile of a distribution is the value with p percent of the observations less than it.

6 7

7 2334

7 5777899

8 00123334

8 569

9 03

Jenny earned a score of 86 on her test. How did she perform relative to the rest of the class?

Her score was greater than 21 of the 25 observations. Since 21 of the 25, or 84%, of the scores are below hers, Jenny is at the 84th percentile in the class’s test score distribution.

6 7

7 2334

7 5777899

8 00123334

8 569

9 03

+ TIPs!!!!!!

Percentiles should be whole numbers.

If you get a decimal, round the answer to the nearest integer

If you get 0.9213, assume 92nd percentile.

If two observations have the same value, they will be at the same percentile. To find the percentile, calculate the percent of the values in the distribution that are below both values

Unfortunately , there is no universally agreed upon definition of “percentiles”

On the AP exam, students will be allowed to use any reasonable definition of “percentile” on the free response section.

+ Your Turn: Your Turn: Wins in Major League Baseball

The stemplot below shows the number of wins for each of the 30 Major League Baseball teams in 2009.

Find the percentiles for the following teams:

(a) The Colorado Rockies, who won 92 games.

(b) The New York Yankees, who won 103 games.

(c) The Kansas City Royals and Cleveland Indians, who both won 65 games.

5 9 6 2455 7 00455589 8 0345667778 9 123557 10 3

Key: 5|9 represents a team with 59 wins.

+ Solution:

(a) The Colorado Rockies, who won 92 games, are at the 80th percentile, since 24/30 teams had fewer wins than they did.

(b) The New York Yankees, who won 103 games, are at about the 97th percentile since 29/30 teams had fewer wins than they did.

(c) The two teams with 65 wins, the Kansas City Royals and the Cleveland Indians, are at the 10th percentile since only 3/30 teams had fewer wins than they did.

+ Cumulative Relative Frequency Graphs

A cumulative relative frequency graph (or ogive) displays the cumulative relative frequency of each class of a frequency distribution.

Age of First 44 Presidents When They Were Inaugurated

Age Frequency

Relative frequency

Cumulative frequency

Cumulative relative frequency

40-44

2 2/44 = 4.5%

2 2/44 = 4.5%

45-49

7 7/44 = 15.9%

9 9/44 = 20.5%

50-54

13 13/44 = 29.5%

22 22/44 = 50.0%

55-59

12 12/44 = 34%

34 34/44 = 77.3%

60-64

7 7/44 = 15.9%

41 41/44 = 93.2%

65-69

3 3/44 = 6.8%

44 44/44 = 100%

+

Use the graph from page 88 to answer the following questions.

Was Barack Obama, who was inaugurated at age 47, unusually young?

Estimate and interpret the 65th percentile of the distribution

Interpreting Cumulative Relative Frequency Graphs

47

11

65

58

+ Your turnYour turn: : State Median Household IncomesHere is a table showing the distribution of median household incomes for the 50 states and the District of Columbia.

MedianIncome($1000s)

FrequencyRelative

FrequencyCumulativeFrequency

CumulativeRelative

Frequency

35 to < 40 1

40 to < 45 10

45 to < 50 14

50 to < 55 12

55 to < 60 5

60 to < 65 6

65 to < 70 3

+

MedianIncome($1000s)

FrequencyRelative

FrequencyCumulativeFrequency

CumulativeRelative

Frequency

35 to < 40 1 1/51 = 0.020 1 1/51 = 0.020

40 to < 45 10 10/51 = 0.196 11 11/51 = 0.216

45 to < 50 14 14/51 = 0.275 25 25/51 = 0.490

50 to < 55 12 12/51 = 0.236 37 37/51 = 0.725

55 to < 60 5 5/51 = 0.098 42 42/51 = 0.824

60 to < 65 6 6/51 = 0.118 48 48/51 = 0.941

65 to < 70 3 3/51 = 0.059 51 51/51 = 1.000

+ Make a Cumulative Frequency Graph: OGIVE

What does the point ( 50, 0.49) and

( 50,0.725) mean?The point at (50,0.49) means 49% of the states had median household incomes less than $50,000.

The point at (55, 0.725) means that 72.5% of the states had median household incomes less than $55,000.

Thus, 72.5% - 49% = 23.5% of the states had median household incomes between $50,000 and $55,000 since the cumulative relative frequency increased by 0.235.

Due to rounding error, this value is slightly different than the relative frequency for the 50 to <55 category.

+ 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

+ Measuring Position: z-Scores

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

+ We can see that Jenny’s score of 86 is “above average” but how far above is it?

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?

This means that Jenny’s score was almost 1 full standard deviation above the mean.

zx mean

standard deviation

86 80

6.070.99

+Calculating z-scores79 81 80 77 73 83 74 93 78 80 75 67 73

77 83 86 90 79 85 83 89 84 82 77 72

6 | 77 | 23347 | 57778998 | 001233348 | 5699 | 03

6 | 77 | 23347 | 57778998 | 001233348 | 5699 | 03

Jenny: z = (86-80)/6.07 z = 0.99 {above average = +z}

Kevin: z = (72-80)/6.07 z = -1.32 {below average = -z}

Kim: z = (80-80)/6.07 z = 0 {average z = 0}

+ Using z-scores for Comparison

zstats 86 80

6.07

zstats 0.99

zchem 82 76

4

zchem 1.5

Standardized values can be used to compare scores from two different distributions.

Jenny’s Scores:

Statistics Test: mean = 80, std dev = 6.07Chemistry Test: mean = 76, std dev = 4Jenny got an 86 in Statistics and 82 in Chemistry.On which test did she perform better?

Although she had a lower score, she performed relatively better in Chemistry.

+Percentiles

Another measure of relative standing is a percentile rank.

pth percentile: Value with p% of observations below it.

median = 50th percentile {mean=50th %ile if symmetric}

Q1 = 25th percentile

Q3 = 75th percentile6 | 77 | 23347 | 57778998 | 001233348 | 5699 | 03

6 | 77 | 23347 | 57778998 | 001233348 | 5699 | 03

Jenny got an 86.22 of the 25 scores are ≤ 86.Jenny is in the 22/25 = 88th %ile.

+

Example, p. 93

Transforming DataTransforming converts the original observations from the original units of measurements to another scale. Transformations can affect the shape, center, and spread of a distribution.

Adding the same number x (either positive, zero, or negative) to each observation:

•adds x to measures of center and location (mean, median, quartiles, percentiles), but

•Does not change the shape of the distribution or measures of spread (range, IQR, standard deviation).

Effect of Adding (or Subracting) a Constant

n Mean sx Min Q1 M Q3 Max IQR Range

Guess(m) 44 16.02 7.14 8 11 15 17 40 6 32

Error (m) 44 3.02 7.14 -5 -2 2 4 27 6 32

+

Example, p. 95

Transforming Data

Multiplying (or dividing) each observation by the same number b (positive, negative, or zero):

•multiplies (divides) measures of center and location by b

•multiplies (divides) measures of spread by |b|, but

•does not change the shape of the distribution

Effect of Multiplying (or Dividing) by a Constant

n Mean sx Min Q1 M Q3 Max IQR Range

Error(ft) 44 9.91 23.43 -16.4 -6.56 6.56 13.12 88.56 19.68 104.96

Error (m) 44 3.02 7.14 -5 -2 2 4 27 6 32

+ 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.

+ Density Curve

Definition:

A density curve is a curve that•is always on or above the horizontal axis, and•has area underneath is exactly 1 it. ( 100% of observations)

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.

Density Curves

Density Curves come in many different shapes; symmetric, skewed, uniform, etc.The area of a region of a density curve represents the % of observations that fall in that region.The median of a density curve cuts the area in half.The mean of a density curve is its “balance point.”

Area under the curve

+ 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

+Section 2.1Describing Location in a Distribution

In this section, we learned that…

There are two ways of describing an individual’s location within a distribution – the percentile and z-score.

A cumulative relative frequency graph allows us to examine location within a distribution.

It is common to transform data, especially when changing units of measurement. Transforming data can affect the shape, center, and spread of a distribution.

We can sometimes describe the overall pattern of a distribution by a density curve (an idealized description of a distribution that smooths out the irregularities in the actual data).

Summary

+ Lets Practice:

# 1, 4, 6,

9,10, 14,

15,16, 27

+

+

+ 2.4:

Larry’s wife should gently break the news that being in the 90th percentile is not good news in this situation. About 90% of men similar to Larry have lower blood pressures. The doctor was suggesting that Larry take action to lower his blood pressure.

+ 2.6. Peter’s time was slower than 80% of his previous race times that season, but it was slower than only 50% of the racers at the league championship meet.

+

+ DO NOW:

In 2010, Taxi Cabs in New York City charged an initial fee of $2.50 plus $2 per mile. In equation form, fare = 2.50 + 2(miles). At the end of a month a businessman collects all of his taxi cab receipts and calculates some numerical summaries. The mean fare he paid was $15.45 with a standard deviation of $10.20. What are the mean and standard deviation of the lengths of his cab rides in miles?

Answer:Solving the equation fare = 2.50 + 2(miles) for miles, we get

2.50

2

faremiles

. Thus,

to find the mean number of miles, subtract 2.50 from the mean fare to get 12.95 and divide by 2 to get 6.475 miles. To find the standard deviation, we divide the standard deviation of the fares by 2 to get 5.10 miles. We don’t subtract 2.50 since subtracting a constant from each observation does not change the spread of a distribution.

+Looking Ahead…

We’ll learn about one particularly important class of density curves – the Normal Distributions

We’ll learn The 68-95-99.7 RuleThe Standard Normal DistributionNormal Distribution Calculations, andAssessing Normality

In the next Section…

+Chapter 2Modeling Distributions of Data

2.1 Describing Location in a Distribution

2.2 Normal Distributions

+Section 2.2Normal 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

+ Because density curves are idealized

descriptions, we need to distinguish between the mean and standard deviation of the density curve ( sample) vs. the mean and standard deviation from the actual observations. ( Population)

Paramete

rs

(Population)

Statistics

( Sample)

mean

ActualIdeal

st. dev

x

s

+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.

+ 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).

Why Normal Distributions?

1) Normal distributions describe many sets of real data. For example...

Scores on tests taken by many people (SAT's, psychological tests).

Repeated careful measurements of the same quantity.

Characteristics of biological populations (such as yields of corn and lengths of animal pregnancies).

Why Normal Distributions?

2) Normal distributions are good approximations to the results of many kinds of chance outcomes, such as tossing a coin many times.

3) Third, and most important, we will see that many statistical inference procedures based on Normal distributions work well for other roughly symmetric distributions.

Be aware...

Even though many sets of data follow a Normal distribution, many do not.

Income distributions are skewed right. Some symmetric distributions are NOT Normal. Don't assume a distribution is Normal just because it looks like it.

+

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.

The 68-95-99.7 Rule

Example: The batting averages for Major League Baseball players in 2009, the mean of the 432 batting averages was 0.261 with a standard deviation of 0.034.

Suppose that the distribution with

= 0.261 , = 0.034(a) Sketch a Normal density curve for this distribution of batting averages. Label the points that are 1, 2, and 3 standard deviations from the mean.

(b) What percent of the batting averages are above 0.329? Show your work.

(c) What percent of the batting averages are between 0.193 and .295? Show your work.

+ (b) Here is a curve showing the proportion of batting averages above 0.329.

Since 0.329 is exactly two standard deviations above the mean, we know that about 95% of batting averages will be between 0.193 and 0.329.

Since the curve is symmetric, half of the remaining 5%, or 2.5% should be above 0.329.

+ (c) Here is a curve showing the proportion of batting averages between 0.193 and 0.295.

We know that about 68% of batting averages are between 0.227 and 0.295.

Also, we know that half of the difference between 95% and 68% should be between 0.193 and 0.227.

Therefore, about 13.5% + 68% = 81.5% of batting averages will be between 0.193 and 0.295.

+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 -

Because all Normal distributions are the same once we standardize, you can now find percentages under Normal curves without Calculus, using the standard Normal table.

Recall an area under a density curve is a proportion of the observations in a

distribution.

+

+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:

Be aware...

A common mistake is to look up a z-value in Table A and report the entry corresponding to that z-value, regardless of whether the problem asks for the area

to the left or to the right of that z-value.

Always sketch the standard Normal curve, mark the z-value, shade the area of interest, and make

sure your answer is reasonable in the context of the problem.

+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.

Example, p. 117

Can you find the same proportion using a different approach?

1 - (0.1056+0.2090) = 1 – 0.3146

= 0.6854

+

Find the proportion between -1.23 and 2.11

(Area left of 2.11) – (Area right of -1.23)

0.9826 – 0.1093 = 0.8733

Using the Standard Normal Table

Cholesterol in Young Boys

For 14-year-old boys, the mean is µ=170 milligrams of cholesterol per deciliter of blood (mg/dl) and the standard deviation is σ = 30 mg/dl.

Levels above 240 mg/dl may require medical attention. What percent of 14-year-old boys have more than 240 mg/dl of cholesterol?

Draw a picture Use the table

240 1702.33

30( 2.33) 1 0.9901

0.0099

0.99%

z

P z

Cholesterol in Young BoysWhat percent of 14-year-old boys have blood

cholesterol between 170 and 240 mg/dl

Area between 0 and 2.33 =

area below 2.33-area below 0.00

= .9901 - .5000 = .4901

Calculator :

normalcdf ( 170, 240 , 170, 30) = .4901846

lb ub mean sd

+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

+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.

+ FINDING A VALUE WHEN GIVEN A PROPORTION

To find the observed value that corresponds to a given percentile

Rule :

Use Table A backwards.

Find the given proportion in the body of the table and read the corresponding z value from the left column and top row.

+ Cholesterol in boys problem

Mean = 170 mg/dl , Sd = 30 mg/dl

Find the First Quartile of the distribution of blood cholesterol.

Answer: First quartile means area from the left is 0.25.

Look in the body of Table A for number closest to 0.25.

It is 0.2514. The z value for this is z = -0.67 ( which is standardized).

To unstandardize: It helps to find the x value.

Solve for x x = 149.9.

The 1st Quartile of blood cholesterol is about 150 mg/dl.

1700.67

30

x

+ Calculator:

To get area from z value:

2nd Distribution ( Vars)

normalcdf ( lower bound, upper bound, mean, Sd)

To get z value from area (unstandardize)

2nd Distribution ( Vars)

invNorm ( proportion, mean, sd)

+IQ Scores

Based on N(100,16) what % of people’s IQ scores would you expect to be: a) over 80? b) under 90? c) between 112 and 132? d) What IQ represents the 15th percentile e) What IQ represents the 98th percentile f) What is the IQR of the IQ’s

+ AP EXAM COMMON ERROR

You will not get full credit if you use calculator speak.

Look at AP Exam tip P- 124.

Label each number and also draw the sketch.

+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

+ Do Check your understanding: P- 119.

+ P- 119

+ P- 119

+ P- 119

+ P- 119

+ How to use calculator for normality plot.

Make list L1 ready

Then do 2nd Stat plot: Choose : type 6th one.

+ Do: Page : 133 # 63, 66, 69-74( all)

+Section 2.2Normal 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

+Section 2.2Normal 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.

Summary