In Summary:
Sept. 5 Statistic for the day:
The standard deviation is roughly 0.25 times the range of
the middle 95% of the data. The mean is the arithmetic
average.
Mean height of U.S. females / males aged 20-29:
64.3 inches / 69.9 inches
Mean self-reported heights for a PSU sample:
64.4 inches / 70.6 inches
Conversely, if you want to visualize a histogram and you
only know the mean and the standard deviation:
1.  Put the center at the mean.
2.  Go out 2 standard deviations on either side of the mean.
3.  Draw a bell-shaped histogram centered at the mean and
dropping off on either side so that 95% of the area is
between the two S. D. marks from part 2.
Assignment for Friday:
Read pages 150-156 in Chapter 8
Try Exercises 2, 3, 8, and 10 on pages 156-159
More on this “bell-shaped” idea next time!
Smoothing the histogram:
The Normal Curve (Chapter 8)
Percent
10
A histogram tends to be rough. To replace it with a bell
shaped curve:
5
Center the “bell” at the mean.
The “bell” should be just wide enough so that the middle
95% of the bell is 4 standard deviations wide.
0
15
20
25
HandSpan
Range of middle 95% is roughly 24 – 16 = 8
Standard deviation is roughly 8/4 = 2
Standard deviation from formula: 1.927
This makes systematic, accurate predictions of all sorts
possible, provided the bell shape is appropriate for the
underlying population.
Recall this dataset on handspans from the
previous lecture:
Histogram of HandSpan, with Normal Curve
23.5
21.0
21.5
23.0
21.0
22.0
22.5
23.0
21.0
22.5
23.0
22.0
22.5
24.5
24.5
20.5
22.5
22.5
23.0
21.5
22.0
20.5
22.0
24.5
21.5
21.0
23.5
22.0
22.0
22.0
24.0
20.5
24.0
22.0
24.5
22.0
19.0
23.5
20.5
22.0
18.0
24.0
22.5
22.0
20.0
21.5
21.0
23.5
21.5
21.5
22.0
22.5
20.5
21.0
24.0
22.0
Women (n = 89)
21.5
18.0
20.5
18.5
19.0
18.5
20.0
20.0
20.0
21.0
19.5
19.0
16.0
17.5
18.5
20.0
19.0
19.0
20.0
18.0
20.5
21.5
20.5
20.5
19.5
22.0
20.0
21.0
19.5
19.5
18.5
21.5
20.0
22.5
17.0
17.0
20.5
21.0
20.0
23.0
21.5
21.5
20.0
18.5
18.5
18.5
20.0
20.0
20.5
21.0
19.0
20.5
21.0
20.0
21.0
20.0
18.5
16.5
19.0
19.0
19.0
20.5
17.0
19.5
19.0
20.0
19.5
20.5
20.0
19.5
19.0
18.5
20.0
17.0
21.0
18.5
20.5
19.5
19.0
20.5
21.0
18.5
17.5
19.5
18.5
19.5
20.0
20.0
18.5
Frequency
20
Men (n = 78)
23.5
22.5
24.5
24.5
20.0
21.0
24.0
21.0
21.0
22.5
23.0
24.0
24.5
23.0
10
0
15
20
25
HandSpan
Mean = 20.86
Standard deviation = 1.927
1
Histogram of HandSpan, with Normal Curve
Histogram of Height, with Normal Curve
20
Frequency
Frequency
30
10
20
10
0
0
15
20
60
25
Research Question 2: How high should
I build my doorways so that 99% of the
people will not have to duck?
Z-Scores: Measurement in Standard
Deviations
Given the mean (68 inches), the standard deviation (4
inches), and a value (75 inches) compute
75 − mean 75 − 68
Z=
=
= 1.75
SD
4
This says that 75 is 1.75 standard deviations
above the mean.
80
Mean = 68 inches or 5 feet 8 inches
Standard deviation = 4 inches
Mean = 20.86
Standard deviation = 1.927
Histogram of Height, with Normal Curve
30
Frequency
Research Question 1: If I built my doors
75 inches (6 feet 3 inches) high, what
percent of the people would have to
duck?
70
Height
HandSpan
20
10
0
60
70
80
Height
Question 1
(x=75)
Question 2
(x=??)
Q1: The value of x is 75; find the amount of distribution above it.
Q2: Find the value of x so that 99% of the distribution is below it.
Compute your Z-score.
1.  How many standard deviations are you
above or below the mean.
Use:
Mean = 68 inches
Standard deviation = 4 inches
2. Now use the table from the book (p. 157) to
determine what percentile you are.
2
Compare Heights of Females and Males
Stat 100 students Sp01
80
Assume male heights have a normal distribution with
mean 70 inches and st dev 3 inches. Assume female
heights have a normal distribution with mean 64 inches
and st dev 3 inches.
Height
What is your Z-Score within your sex?
70
What is your percentile within your sex?
60
Female
Male
Sex
Histogram of Height, with Normal Curve
Answer to Question 1: What percent of people would
have to duck if I built my doors 75 inches high?
From the standard normal table in the book: .96 or
96% of the distribution is below 1.75. Hence, .04
or 4% is above 1.75.
Frequency
Recall: 75 has a Z-score of 1.75
30
20
4% in here
10
0
60
So 4% of the distribution is above 75 inches.
70
75
80
Height
Question 1
(x=75)
The value at x is 75; find the amount of distribution above
it. Convert 75 to Z = 1.75 and use Table 8.1 on p. 157.
Question 2: What is the value so that 99% of the
distribution is below it? (called the 99th percentile.)
2.  Now convert it over to inches:
2.33 =
h99 − 68
4
h99 = 68 + 2.33(4) = 77.3
Therefore, 99% of the distribution is shorter than 77.3
inches (6 foot 5.3 inches) and that’s how high the door
should be built.
30
Frequency
1.  Look up the Z-score that corresponds to the 99th
percentile. From the table: Z = 2.33.
Histogram of Height, with Normal Curve
20
10
99% in here
0
60
70
80
Height
Question 2
77.3 inches is the 99th percentile
To find the value so that 99% of the distribution is below it: Look up
the Z-score for the 99th percentile and convert it back to inches.
3
Answer these questions:
Research Question 1: What percent of people are less
than six feet (72 inches) tall?
To answer question 1, first convert
72 inches to a z-score:
Given the mean (68 inches), the standard deviation (4
inches), and a value (72 inches), compute
Research Question 2: What is the first quartile of
heights?
(Assume that adults’ heights are normally distributed
with mean 68 inches and standard deviation 4 inches.)
Answer to Question 2: What is the first quartile of heights?
Translation: “First quartile” means 25th percentile, which
means .25 are below that height.
From p. 157: Find the z-score corresponding to the 25th
percentile.
–.67
Now convert this z-score into a height:
Z − score =
h − 68
4
Z = (72-mean) / SD = (72-68) / 4 = 1
This says that 72 is 1 standard deviation above the
mean.
What proportion of heights are below Z-score=1 ? .84 or 84%
Shaquille O’Neal is 7 feet 1 inch or
85 inches tall. How many people in
the country are taller?
We will assume that heights are normally distributed
with mean 68 inches and standard deviation 4 inches.
O’Neal’s Z-score is Z = (85-68)/4 = 4.25. In other
words
O’Neal is 4.25 standard deviations above the mean(!)
h = 68 + 4( Z − score)
There are roughly 305 million people in US.
About 49% are over the age of 20 (Census Bureau).
That is about 150 million.
Hence, there should be roughly
.000011 times 150 million
or 1650 people taller than Shaquille
O’Neal.
Note: This is an extremely rough calculation, since the
normal distribution approximation is less accurate at the
extremes. Also, cutting off at age 20 might miss some
tall teens!
There is only 0.000011 of the normal distribution above
4.25 standard deviations.
Page 157
Determine the percentage of a normal distribution
falling below each of the following standard scores:
•  -1.00
•  1.96
•  0.84
•  Determine the percentage of a normal distribution
falling above each of the following standard scores:
•  1.28
•  -0.25
•  2.33
•
4