1.3

Report
Looking at data: distributions
- Density curves and normal distributions
IPS 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 curves
A 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
Normal (or “Gaussian”) distributions are a family of symmetric, singlepeaked density curves defined by the mean m (mu) and standard
deviation s (sigma). A normal distribution is symbolized by N (m, s).
1
e
2
f ( x) 
1  xm 
 

2 s 
2
x
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…
x
A family of normal curves
Here the means are different
(m = 10, 15, and 20) while the
standard deviations are all the
same (s = 3).
0
2
4
6
8
10
12
14
16
18
Visually, how can you tell all three curves have
the same standard deviation of s = 3 ?
Which of the above corresponds to the normal
curve with mean of m = 10 ?
20
22
24
26
28
30
A family of normal curves
Here the means are the same (m =
15) while the standard deviations
are different (s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Visually, how can you tell that all three curves
have the same mean of m = 15 ?
Which of the above corresponds to the normal
curve with standard deviation of s = 6?
Characteristics of any normal curve
 a normal curve defined by its mean m and std. deviation s
 larger s 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 m − s and m + s
 total area beneath equals 1
 area on either side of mean is .50 (due to symmetry)
 area beneath represents probability
The 68-95-99.7 Rule for normal distributions

About 68% of all observations are
within 1 standard deviation
of the mean.
Inflection point
64.5 ± 1(2.5) → 64.5 ± 2.5
→ 62 to 67 inches

About 95% of all observations are
within 2 s 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 s of the mean.
64.5 ± 3(2.5) → 64.5 ± 7.5
mean µ = 64.5
standard deviation s = 2.5
N(µ, s) = N(64.5, 2.5)
→ 57 to 72 inches
Reminder: µ (mu) is the mean of the idealized normal curve, while x 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.
The standard normal distribution
Because all normal distributions share the same properties, we can
standardize our data to transform any normal curve N (m, s) into the
standard normal curve N (0,1).
N(64.5, 2.5)
N(0,1)
=>
x
Standardized height (no units)
For any x we can calculate a “standardized value” z (called a z-score).
z
Standardizing: calculating z-scores
How tall is X=67 inches compared to the mean of 64.5?
Clearly it is 2.5 inches above average. (X−m = 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 m.
z
(x  m )
s
 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.
Example: Women heights
N(µ, s) =
N(64.5, 2.5)
Women’s heights follow the N(64.5″,2.5″)
distribution. What percent of women are
Area=
Area= .50
???
Area
Area == ???
.34
shorter than 67 inches tall (that’s 5′7″)?
mean µ = 64.5"
standard deviation s = 2.5"
X (height) = 67"
m = 64.5″ X = 67″
z =0
z =1
We calculate z, the standardized value of x:
z
(x  m)
s
, z
(67  64.5) 2.5

 1  so 67" is 1 std. dev. above the mean
2.5
2.5
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%.
Using Table A to find percent of women shorter than 67”
The area under the
standard normal curve to
the left of z =1.00 is 0.8413.
N(µ, s) =
N(64.5”, 2.5”)
Area ≈ 0.84
Conclusion:
Area ≈ 0.16
84.13% of women are shorter than 67″.
(By subtraction, 1 − 0.8413, or 15.87%, of
women are taller than 67".)
m = 64.5” x = 67”
z=1
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)?
x  820
m  1026
s  209
(x  m)
z
s
(820 1026)
209
 206
z
 0.99
209
T able A : area t o
z
t heleft of
z  .99 is 0.1611
or approx.16%.
Area right of 820
=
=
= .8389
Total area
1
≈ 84%
−
−
Area left of 820
0.1611
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.
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.
c) subtract the smaller
area from the larger area.
area between z1 and z2 =
A common mistake is to
subtract the z-scores; only
subtract areas.
(area left of z1) – (area left of z2)
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?
x  720
m  1026
s  209
(x  m)
z
s
(720 1026)
209
 306
z
 1.46
209
T able A : area under
z
N(0,1)t o t heleft of
z   1.46 is 0.0721
or approx.7%.
Area in between
=
=
=
Area left of 820
0.1611
.8900 ≈ 9%
−
−
Area left of 720
0.0721
About 9% of all students who take the SAT have scores
between 720 and 820.
Comparing Apples to Oranges
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.
(x  m )
z
s
N(0,1)
Who is “taller”? A 67-inch woman or a 71-inch man?
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.
z
(x  m)
s
, z
(67  64.5) 2.5

1
2.5
2.5
How many std. deviations above the mean is 71” man? (compared to the
male mean)
z
(x  m)
s
, z
(71  69)
2

 .8
2.5
2.5
Example: Gestation time in malnourished mothers
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.
What improvement did we get
by adding better food?
m 266
s 15
m 250
s 20
180
200
220
240
260
280
Gestation time (days)
Vitamins only
Vitamins and better food
300
320
Under each treatment, what percent of mothers failed to carry their babies at
least 240 days?
Vitamins only
m = 250, s = 20,
x = 240
x  240
m  250
s  20
z
(x  m)
s
(240  250)
20
170
 10
z
 0 .5
20
(half a standarddeviation)
T able A : area under N(0,1)t o
t heleft of z  0.5 is 0.3085.
z
190
210
230
250
270
290
Gestation time (days)
Vitamins only: 30.85% of women
would be expected to have gestation
times shorter than 240 days.
310
Vitamins and better food
m = 266, s = 15,
x = 240
x  240
m  266
s  15
(x  m)
z
s
(240  266)
z
15
 26
206
z
 1.73
15
(almost 2 sd from mean)
T able A : area under N(0,1)t o
t heleft of z - 1.73is 0.0418.
221
236
251
266
281
296
311
Gestation time (days)
Vitamins and better food: 4.18% of women
would be expected to have gestation
times shorter than 240 days.
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%).
Determining normal percentiles
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.
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 m, s
and Z and then solve for X.
z
(x  m)
s
Example: Women’s heights
Women’s heights follow the N(64.5″,2.5″)
distribution. What is the 25th percentile for
women’s heights?
mean µ = 64.5"
standard deviation s = 2.5"
proportion = area under curve = 0.25
z
(x  m)
s
On the first page of Table A, a left-hand area of 0.25 corresponds to
z = −0.67 (closest area is .2514).
z
(x  m)
( x  64.5)
 .67 
2 .5
s
−.67(2.5) = X − 64.5
or
X = 64.5 −.67(2.5) = 62.825
The 25th percentile for women’s heights is roughly 62.8”.
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.

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