Looking at data: distributions- Density curves and normal distributionsIPS section 1.3

© 2006 W.H. Freeman and Company (authored by Brigitte Baldi, University of California-Irvine; adapted by Jim Brumbaugh-Smith, Manchester College)

Objectives

Density Curves and Normal Distributions

Understand interpretation of density curves

Understand characteristics of normal distributions

Apply the 68-95-99.7 Rule

Compute and interpret a z-score for normal data

Calculate normal probabilities

Calculate normal percentiles

Compare values from different normal populations

Assess normality of data using normal quantile plots

Terminology

Density curve (or “probability density function”)

Normal distributions (or “normal curves”

The “68-95-99.7 Rule” (or the “Empirical Rule”)

Standard normal distribution

Standardized value (or “z-score”)

Normal quantile plot

Density curvesA density curve is a mathematical model of a distribution. It can be thought of as a smoothed version of the underlying histogram.

It is always on or above the horizontal axis.

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 within that range.

Histogram of sample data with theoretical density curve

describing the population

Density curves can come in any

imaginable shape (as long as the area

beneath equals 1).

Some are well-known mathematically and

others aren’t. As with histograms, peaks

represent ranges with frequently occurring

values while valleys and tails correspond

to less frequently occurring values.

Normal distributions

While not essential to know, the above equation represents the red normal curve.

e = 2.71828… (the base of the natural logarithm)

π = pi = 3.14159…

Normal (or “Gaussian”) distributions are a family of symmetric, single-

peaked density curves defined by the mean (mu) and standard

deviation (sigma). A normal distribution is symbolized by N ().

2

2

1

2

1)(

x

exf

xx

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

A family of normal curves

Here the means are different

( = 10, 15, and 20) while the

standard deviations are all the

same ( = 3).

Visually, how can you tell all three curves have the same standard deviation of = 3 ?

Which of the above corresponds to the normal curve with mean of = 10 ?

A family of normal curves

Here the means are the same ( =

15) while the standard deviations

are different ( = 2, 4, and 6).

Visually, how can you tell that all three curves have the same mean of = 15 ?

Which of the above corresponds to the normal curve with standard deviation of = 6?

Characteristics of any normal curve

a normal curve defined by its mean and std. deviation

larger implies more variation in the data, so the normal curve will be broader

exactly symmetric and single peaked

in theory, no lower or upper limit on data represented

curvature changes at − and +

total area beneath equals 1

area on either side of mean is .50 (due to symmetry)

area beneath represents probability

mean µ = 64.5 standard deviation = 2.5

N(µ, ) = N(64.5, 2.5)

The 68-95-99.7 Rule for normal distributions

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

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

About 68% of all observations are within

1 standard deviation

of the mean.

64.5 ± 1(2.5) → 64.5 ± 2.5

→ 62 to 67 inches

About 95% of all observations are within

2 of the mean.

64.5 ± 2(2.5) → 64.5 ± 5

→ 59.5 to 69.5 inches

Almost all (99.7%) observations are

within 3 of the mean.

64.5 ± 3(2.5) → 64.5 ± 7.5

→ 57 to 72 inches

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 any x we can calculate a “standardized value” z (called a z-score).

N(0,1)

=>

z

x

N(64.5, 2.5)

Standardized height (no units)

z (x )

How tall is X=67 inches compared to the mean of 64.5?

Clearly it is 2.5 inches above average. (X− = 67−64.5 = 2.5)

But how far is it above the mean relative to the standard deviation?

A z-score measures the number of standard deviations that a data

value x is from the mean .

Standardizing: calculating z-scores

When X is larger than the mean, z is positive.

When X is smaller than the mean, z is negative.

For the mean itself, z=0.

mean µ = 64.5"

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

We calculate z, the standardized value of x:

mean theabove dev. std. 1 is 67" so 15.2

5.2

5.2

)5.6467( ,

)(

z

xz

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

than 67″ is approximately, .50 + (half of .68) = .50 + .34 = .84, or 84%.

Area= ???

Area = ???

N(µ, ) = N(64.5, 2.5)

= 64.5″ X = 67″

z = 0 z = 1

Example: Women heights

Women’s heights follow the N(64.5″,2.5″)

distribution. What percent of women are

shorter than 67 inches tall (that’s 5′7″)?

Area= .50

Area = .34

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".)

The area under the

standard normal curve to

the left of z =1.00 is 0.8413.

Using Table A to find percent of women shorter than 67”

Using Table A

(…)

Table A gives the area under the standard normal curve to the left of any z-score.

.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

Using Table A to find right-hand areas

Since the total area is 1,

right-hand area = 1 − left-hand area

area right of z = 1 − area left of z

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

toarea :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 considered to be zero for a normal distribution due to the idealized smoothing of density curves.

Area right of 820 = Total area − Area left of 820= 1 − 0.1611= .8389 ≈ 84%

Using Table A to find in-between areas

To calculate the area between two X values:

a) compute their z-scores separately.

b) find the area to the left

of each z-score.

area between z1 and z2 =

(area left of z1) – (area left of z2)A common mistake is to

subtract the z-scores; only

subtract areas.

c) subtract the smaller

area from the larger area.

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

scholarship but not to compete) as a combined SAT score of 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 46.1 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 in between = Area left of 820 − Area left of 720 = 0.1611 − 0.0721

= .8900 ≈ 9%

N(0,1)

z (x )

When working with normally distributed data we can manipulate it and then compare seemingly non-comparable distributions.

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

Comparing Apples to Oranges

Who is “taller”? A 67-inch woman or a 71-inch man?

15.2

5.2

5.2

)5.6467( ,

)(

z

xz

Women’s Heights ~ N(64.5”, 2.5”)

Men’s Heights ~ N(69”, 2.5”)

We already saw a 67” women is 1 std. deviation above the mean for women.

How many std. deviations above the mean is 71” man? (compared to the

male mean)

8.5.2

2

5.2

)6971( ,

)(

z

xz

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

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

Example: 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%).

Find the height (X) such that 25% of women are below this height.

That is, find the 25th percentile for women’s heights.

When you know the desired percentage, but you don’t know the X

value that represents the cut-off, you need to use Table A backward.

Determining normal percentiles

1. Find the percentile (left-hand area) in the body of Table A.

2. Find the corresponding z-score in the margin of Table A.

3. Transform Z back to the X scale (that is, inches). Plug in , and Z and then solve for X.

)(

x

z

mean µ = 64.5"

standard deviation = 2.5" proportion = area under curve = 0.25

On the first page of Table A, a left-hand area of 0.25 corresponds to z = −0.67 (closest area is .2514).

The 25th percentile for women’s heights is roughly 62.8”.

Example: Women’s heights

5.2

)5.64(67.

x

)(

x

z

Women’s heights follow the N(64.5″,2.5″)

distribution. What is the 25th percentile for

women’s heights?

)(

x

z

−.67(2.5) = X − 64.5 or X = 64.5 −.67(2.5) = 62.825

Assessing normality

In problem solving we often assume data is normal.

In statistical work we often need to verify that data is close to normal.

Is the data in the following histogram fairly normal?

The overall shape is fairly symmetric with one main peak in the center.

Recall that histograms can be misleading depending on the number of vertical bars used.

Adding the best fitting normal curve N(100, 9.7) suggests a fairly good fit.

Normal quantile plots

A normal quantile plot (Q-Q plot in SPSS) is a plot of z-scores

versus the actual data (X). If the data is close to normal than the

quantile plot will be close to linear.

Here the plot very closely follows the diagonal reference line indicating the data is close to normal.

We generally ignore any slight irregularities at the ends.

Points off to the far left or right indicate outliers.

More quantile plots – skewed data

Quantile plots for skewed data show a sweeping curve.

Concave down → skewed right

Concave up → skewed left

These descriptions apply to SPSS. In figures from our textbook (IPS) the direction of the skewness is the opposite.

More quantile plots – symmetric, non-normal data

Quantile plots for symmetric, but non-normal data show an S shape.

Backwards S → symmetric but with shorter tails than normal

Regular S → symmetric but with longer tails than normal

These descriptions apply to SPSS. In figures from our textbook (IPS) the length of the tails is the opposite.