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Chapter 3: The “Normal” Distributions
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Ch3 exercises: 3.2, 3.3, 3.4, 3.10, 3.26, 3.27, 3.43

Density curves

Mean, median, and standard deviation of a density curve

The Normal distributions
Standardization
The standard normal distribution
Normal distribution calculations
The family of normal distributions plays a central role in statistics. The purpose of this chapter is
to introduce the idea of a density curve;
to explain some of the key properties of the normal family of density curves;
and to show you how to work with the normal distribution table and calculations.
Chapter 3 Concepts
2

Density Curves

Normal Distributions

The 68-95-99.7 Rule

The Standard Normal Distribution

Finding Normal Proportions

Using the Standard Normal Table

Finding a Value When Given a Proportion
Density Curves
4
In Chapters 1 and 2, we developed a kit of graphical and numerical tools
for describing distributions. Now, we’ll add one more step to the strategy.
Exploring Quantitative Data
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.
4. Sometimes the overall pattern of a large number of
observations is so regular that we can describe it by a
smooth curve.
Density Curves
5
Example: Here is a histogram
of vocabulary scores of 947
seventh graders.
The smooth curve drawn over
the histogram is a
mathematical model for the
distribution.
Density Curves
6
The areas of the shaded bars in this
histogram represent the proportion of
scores in the observed data that are
less than or equal to 6.0. This
proportion is equal to 0.303.
Now the area under the smooth curve
to the left of 6.0 is shaded. If the scale
is adjusted so the total area under the
curve is exactly 1, then this curve is
called a density curve. The proportion
of the area to the left of 6.0 is now
equal to 0.293.
Density Curves
7
A density curve is a curve that:
• is always on or above the horizontal axis
• has an area of exactly 1 underneath it
A density curve describes the overall pattern of a
distribution. The area under the curve and above any
range of values on the horizontal axis is the
proportion of all observations that fall in that range.
Density Curves
8

Our measures of center and spread apply to density curves as well as
to actual sets of observations.
•
•
•
Distinguishing the Median and Mean of a Density Curve
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.
8
Density Curves
9

The mean and standard deviation computed from
actual observations (data) are denoted by x and s,
respectively.

The mean and standard deviation of the actual
distribution represented by the density curve are
denoted by µ (“mu”) and  (“sigma”), respectively.
Normal Distributions
10

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.
Normal Distributions
11
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).
The 68-95-99.7 Rule
12
The 68-95-99.7 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.
Normal Distributions
13

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).

Sketch the Normal density curve for this distribution.

What percent of ITBS vocabulary scores are less than 3.74?

What percent of the scores are between 5.29 and 9.94?
,
>i flout J..s'S.
of scores ' "
,,.t111n
,...
(GSS
t1!1
3.11
>.:1.!1
6.11
1.3!1
!1.!1-f
WI
11.+!1
;{.1~
IT8Sscort:
3.f'f
S".'-J
6.11
1.3~
Uf
3.1+
11.1!1
ITBS!.Cbrt:
X- N(6.84, 1.55)
• What percent ofiTBS vocabulary scores are less than 3.74?
z = (x-mu)/sigma =
(3.74- 6.84)/1.55 = -2
P(z < -2) = 0.0228 (or "about 2.5%")
•
What percent ofthe scores are between 3.74 and 9.94?
P( 3.74<x<9.94)
= P((3. 74-6.84)/1.55<Z<(9.94-6.84)/1.55)
= P(-2<Z<2)
= P(Z<2)-P(Z<-2) = 0.9772 - 0.0228 ~ 0.9544
S".'-!1
6.11
IT8S SCbrt:
1.3!1
!l.!l.f
11.-t!l
,
'ftbouus'S
of scores (Ire
tesstktn
,,1-t.
t1!J
3.7-f
S'.j.!J
6.11
1.3!1
!l.!l'f
~.1!f
11.'f!1
,.1!1
ITBSscorc
3.'H
S:'-!1
6.1'1
1.3!1
!1.1'1-
11.'1!1
IT8Sscorc
x ~ N(6.84, 1.55)
• What percent ofiTBS vocabulary scores are less than 3.74?
z = (x-mu)/sigma =
(3.74- 6.84)/1.55 = -2
P(z < -2) = 0.0228 (or "about 2.5%")
•
What percent ofthe scores are between 3.74 and 9.94?
P( 3.74<x<9.94)
= P((3. 74-6.84)/1.55<Z<(9.94-6.84)/1.55)
=P(-2<Z<2)
= P(Z<2)-P(Z<-2) = 0.9772- 0.0228 = 0.9544
..
•
What percent of scores are between 9.94 and 8.39
Z= ((9.94-6.84)/1.55) = 2
z =((8.39-6.84)/1.55) = 1
P( 1 < Z < 2) = 0.9772-0.8413 = 0.1359
(0.9544997-0.6826895)/2 # = 0.1359051
p( 3.74< X< 9.94)
= p((3.74-6.84)/1.55 < z < ((9.94-6.84)/1.55)
= p(-2 < z < 2) = 0.9772 - 0.0228 = 0.9544
p( 5.29 < X< 8.39)
p ((5.29-6.84)/1.55 < z <(8.39-6.84)/1.55)
p(-1 < z < 1) = 0.8413- 0.1587 = 0.6826
(0.9544-0.6826)/2 = 0.1359
3.1'f
S'.7.5
6.14
ITBSscorc
1.,!1
5.!1-f.
1H!1
x ~ N(6.84, 1.55)
 What percent of scores are between 5.29 and 9.94
P(5.29<X<9.94)
=P( (5.29-6.84)/1.55<Z<(9.94-6.84)/1.55)
=P(-1<Z<2)
=P(Z<2)-P(Z<-1) = 0. 9772 - 0.1587
= 0.8185
The Standard Normal Distribution
14

All Normal distributions are the same if we measure in units of size σ
from the mean µ as center.
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
z
x-μ

has the standard Normal distribution, N(0,1).
Key
20|3 means
203 pounds
Stems = 10’s
Leaves = 1’s
The Standard Normal Table
15

Because all Normal distributions are the same when we standardize,
we can find areas under any Normal curve from a single table.
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.
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:
Z
.00
.01
.02
0.7
.7580
.7611
.7642
0.8
.7881
.7910
.7939
0.9
.8159
.8186
.8212
P(z < 0.81) = .7910
Normal Calculations
16
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
Normal Calculations
17
How to Solve Problems Involving Normal Distributions
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.
Normal Calculations
18


According to the Health and Nutrition Examination Study
of 1976-1980, the heights (in inches) of adult men aged
18-24 are N(70, 2.8).
How tall must a man be in the lower 10% for men
aged 18 to 24?
N(70, 2.8)
.10
?
70
Normal Calculations
19

N(70, 2.8)
How tall must a man be in the lower
10% for men aged 18 to 24?
.10
?
70
Look up the closest
probability (closest to 0.10) in
the table.
Find the corresponding
standardized score.
The value you seek is that
many standard deviations
from the mean.
Z = -1.28
Normal Calculations
20

N(70, 2.8)
How tall must a man be in the lower
10% for men aged 18 to 24?
Z = -1.28
.10
?
70
We need to “unstandardize” the z-score to find the observed value (x):
z
x

x    z
x = 70 + z(2.8)
= 70 + [(1.28 )  (2.8)]
= 70 + (3.58) = 66.42
A man would have to be approximately 66.42 inches tall or less to place in
the lower 10% of all men in the population.
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TAB LES
user-F452
•
Table entry for z is the area under
the standard Normal curve to the
left of z.
Table entry
z
TABLE A
Standard Normal cumulative proportions
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
⫺3.4
⫺3.3
⫺3.2
⫺3.1
⫺3.0
⫺2.9
⫺2.8
⫺2.7
⫺2.6
⫺2.5
⫺2.4
⫺2.3
⫺2.2
⫺2.1
⫺2.0
⫺1.9
⫺1.8
⫺1.7
⫺1.6
⫺1.5
⫺1.4
⫺1.3
⫺1.2
⫺1.1
⫺1.0
⫺0.9
⫺0.8
⫺0.7
⫺0.6
⫺0.5
⫺0.4
⫺0.3
⫺0.2
⫺0.1
⫺0.0
.0003
.0005
.0007
.0010
.0013
.0019
.0026
.0035
.0047
.0062
.0082
.0107
.0139
.0179
.0228
.0287
.0359
.0446
.0548
.0668
.0808
.0968
.1151
.1357
.1587
.1841
.2119
.2420
.2743
.3085
.3446
.3821
.4207
.4602
.5000
.0003
.0005
.0007
.0009
.0013
.0018
.0025
.0034
.0045
.0060
.0080
.0104
.0136
.0174
.0222
.0281
.0351
.0436
.0537
.0655
.0793
.0951
.1131
.1335
.1562
.1814
.2090
.2389
.2709
.3050
.3409
.3783
.4168
.4562
.4960
.0003
.0005
.0006
.0009
.0013
.0018
.0024
.0033
.0044
.0059
.0078
.0102
.0132
.0170
.0217
.0274
.0344
.0427
.0526
.0643
.0778
.0934
.1112
.1314
.1539
.1788
.2061
.2358
.2676
.3015
.3372
.3745
.4129
.4522
.4920
.0003
.0004
.0006
.0009
.0012
.0017
.0023
.0032
.0043
.0057
.0075
.0099
.0129
.0166
.0212
.0268
.0336
.0418
.0516
.0630
.0764
.0918
.1093
.1292
.1515
.1762
.2033
.2327
.2643
.2981
.3336
.3707
.4090
.4483
.4880
.0003
.0004
.0006
.0008
.0012
.0016
.0023
.0031
.0041
.0055
.0073
.0096
.0125
.0162
.0207
.0262
.0329
.0409
.0505
.0618
.0749
.0901
.1075
.1271
.1492
.1736
.2005
.2296
.2611
.2946
.3300
.3669
.4052
.4443
.4840
.0003
.0004
.0006
.0008
.0011
.0016
.0022
.0030
.0040
.0054
.0071
.0094
.0122
.0158
.0202
.0256
.0322
.0401
.0495
.0606
.0735
.0885
.1056
.1251
.1469
.1711
.1977
.2266
.2578
.2912
.3264
.3632
.4013
.4404
.4801
.0003
.0004
.0006
.0008
.0011
.0015
.0021
.0029
.0039
.0052
.0069
.0091
.0119
.0154
.0197
.0250
.0314
.0392
.0485
.0594
.0721
.0869
.1038
.1230
.1446
.1685
.1949
.2236
.2546
.2877
.3228
.3594
.3974
.4364
.4761
.0003
.0004
.0005
.0008
.0011
.0015
.0021
.0028
.0038
.0051
.0068
.0089
.0116
.0150
.0192
.0244
.0307
.0384
.0475
.0582
.0708
.0853
.1020
.1210
.1423
.1660
.1922
.2206
.2514
.2843
.3192
.3557
.3936
.4325
.4721
.0003
.0004
.0005
.0007
.0010
.0014
.0020
.0027
.0037
.0049
.0066
.0087
.0113
.0146
.0188
.0239
.0301
.0375
.0465
.0571
.0694
.0838
.1003
.1190
.1401
.1635
.1894
.2177
.2483
.2810
.3156
.3520
.3897
.4286
.4681
.0002
.0003
.0005
.0007
.0010
.0014
.0019
.0026
.0036
.0048
.0064
.0084
.0110
.0143
.0183
.0233
.0294
.0367
.0455
.0559
.0681
.0823
.0985
.1170
.1379
.1611
.1867
.2148
.2451
.2776
.3121
.3483
.3859
.4247
.4641
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TABLE S
67 7
Table entry
Table entry for z is the area under
the standard Normal curve to the
left of z.
z
TABLE A
Standard Normal cumulative proportions (continued)
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
.5000
.5398
.5793
.6179
.6554
.6915
.7257
.7580
.7881
.8159
.8413
.8643
.8849
.9032
.9192
.9332
.9452
.9554
.9641
.9713
.9772
.9821
.9861
.9893
.9918
.9938
.9953
.9965
.9974
.9981
.9987
.9990
.9993
.9995
.9997
.5040
.5438
.5832
.6217
.6591
.6950
.7291
.7611
.7910
.8186
.8438
.8665
.8869
.9049
.9207
.9345
.9463
.9564
.9649
.9719
.9778
.9826
.9864
.9896
.9920
.9940
.9955
.9966
.9975
.9982
.9987
.9991
.9993
.9995
.9997
.5080
.5478
.5871
.6255
.6628
.6985
.7324
.7642
.7939
.8212
.8461
.8686
.8888
.9066
.9222
.9357
.9474
.9573
.9656
.9726
.9783
.9830
.9868
.9898
.9922
.9941
.9956
.9967
.9976
.9982
.9987
.9991
.9994
.9995
.9997
.5120
.5517
.5910
.6293
.6664
.7019
.7357
.7673
.7967
.8238
.8485
.8708
.8907
.9082
.9236
.9370
.9484
.9582
.9664
.9732
.9788
.9834
.9871
.9901
.9925
.9943
.9957
.9968
.9977
.9983
.9988
.9991
.9994
.9996
.9997
.5160
.5557
.5948
.6331
.6700
.7054
.7389
.7704
.7995
.8264
.8508
.8729
.8925
.9099
.9251
.9382
.9495
.9591
.9671
.9738
.9793
.9838
.9875
.9904
.9927
.9945
.9959
.9969
.9977
.9984
.9988
.9992
.9994
.9996
.9997
.5199
.5596
.5987
.6368
.6736
.7088
.7422
.7734
.8023
.8289
.8531
.8749
.8944
.9115
.9265
.9394
.9505
.9599
.9678
.9744
.9798
.9842
.9878
.9906
.9929
.9946
.9960
.9970
.9978
.9984
.9989
.9992
.9994
.9996
.9997
.5239
.5636
.6026
.6406
.6772
.7123
.7454
.7764
.8051
.8315
.8554
.8770
.8962
.9131
.9279
.9406
.9515
.9608
.9686
.9750
.9803
.9846
.9881
.9909
.9931
.9948
.9961
.9971
.9979
.9985
.9989
.9992
.9994
.9996
.9997
.5279
.5675
.6064
.6443
.6808
.7157
.7486
.7794
.8078
.8340
.8577
.8790
.8980
.9147
.9292
.9418
.9525
.9616
.9693
.9756
.9808
.9850
.9884
.9911
.9932
.9949
.9962
.9972
.9979
.9985
.9989
.9992
.9995
.9996
.9997
.5319
.5714
.6103
.6480
.6844
.7190
.7517
.7823
.8106
.8365
.8599
.8810
.8997
.9162
.9306
.9429
.9535
.9625
.9699
.9761
.9812
.9854
.9887
.9913
.9934
.9951
.9963
.9973
.9980
.9986
.9990
.9993
.9995
.9996
.9997
.5359
.5753
.6141
.6517
.6879
.7224
.7549
.7852
.8133
.8389
.8621
.8830
.9015
.9177
.9319
.9441
.9545
.9633
.9706
.9767
.9817
.9857
.9890
.9916
.9936
.9952
.9964
.9974
.9981
.9986
.9990
.9993
.9995
.9997
.9998
EXAMPLE 3.2 Iowa Test scores
the distribution of Iowa Test vocabulary scores for seventh-grade students is close to
Normal.
Suppose that the distribution is exactly Normal with mean μ = 6.84 and standard
deviation σ = 1.55. (These are the mean and standard deviation of the 947 actual scores.)
FIGURE 3.10
The 68–95–99.7 rule applied to the distribution of Iowa Test scores for seventh-grade
students in Gary, Indiana, for Example 3.2. The mean and standard deviation are μ = 6.84
and σ = 1.55.
Figure 3.10 applies the 68–95–99.7 rule to the Iowa Test scores.
The 95 part of the rule says that 95% of all scores are between
μ − 2σ = 6.84 − (2)(1.55) = 6.84 − 3.10 = 3.74
and
μ + 2σ = 6.84 + (2)(1.55) = 6.84 + 3.10 = 9.94
The other 5% of scores are outside this range. Because Normal distributions are
symmetric, half of these scores are lower than 3.74 and half are higher than 9.94.
That is, 2.5% of the scores are below 3.74 and 2.5% are above 9.94.
EXAMPLE 3.3 Iowa Test scores
Mean = 6.84
standard deviation = 1.55
A score of 5.29 is one standard deviation (6.84-5.29 = 1.55) below the mean. What
percent of scores are higher than 5.29? Find the answer by adding areas in the figure.
Here is the calculation in pictures:
Be sure you understand where the 16% came from. We know that 68% of scores are
between 5.29 and 8.39, so 32% of scores are outside that range. These are equally split
between the two tails, 16% below 5.29 and 16% above 8.39.
Z = (5.29 – 6.84) / 1.55 = - 1
1- P(z < -1) = 1- 0.1587 = 0.8413
EXAMPLE 3.4
Example 3.4 Standardizing women’s heights
The heights of women aged 20 to 29 are approximately Normal with μ = 64.3 inches and
σ = 2.7 inches.
The standardized height is
A woman’s standardized height is the number of standard deviations by which her height
differs from the mean height of all young women.
A woman 70 inches tall, for example, has standardized height
or 2.11 standard deviations above the mean.
Similarly, a woman 5 feet (60 inches) tall has standardized height
or 1.59 standard deviations less than the mean height.
Example 3.5
Who qualifies for college sports?
The National Collegiate Athletic Association (NCAA) uses a sliding scale for eligibility
for Division I athletes.
Those students with a 2.5 high school GPA must have a combined score of at least 820
on the Mathematics and Reading parts of the SAT in order to compete in their first
college year.
The scores of the 1.5 million high school seniors taking the SAT this year are
approximately Normal with mean 1026 and standard deviation 209.
What percent of high school seniors meet this SAT requirement of a combined score of
820 or better?
Here is the calculation in a picture: the proportion of scores above 820 is the area under
the curve to the right of 820. That’s the total area under the curve (which is always 1)
minus the cumulative proportion up to 820.
pnorm((820-1026)/209) # = 0.1621534
1-pnorm((820-1026)/209) # = 0.8378466
Z = (820 – 1026) / 209 = -0.986
P(z> - 0.986) = 1 – P(z<-0.986) = 1 - 0.16 = 0.84
About 84% of all high school seniors meet this SAT requirement of a combined math and
reading score of 820 or higher.
EXAMPLE 3.6 The standard Normal table
What proportion of observations on a standard Normal variable z take values less than
1.47?
Solution: To find the area to the left of 1.47, locate 1.4 in the left-hand column of Table
A, then locate the remaining digit 7 as .07 in the top row.
The entry opposite 1.4 and under .07 is 0.9292. This is the cumulative proportion we seek.
Figure 3.11 illustrates this area.
EXAMPLE 3.7 Who qualifies for college sports?
Scores of high school seniors on the SAT follow the Normal distribution with
mean μ = 1026 and standard deviation σ = 209.
What proportion of seniors score at least 820?
Step 1. Draw a picture. The picture shows that area
to the right of 820 = 1 − area to the left of 820
Step 2. Standardize. Call the SAT score x. Subtract the mean and then divide by the
standard deviation to transform the problem about x into a problem about a standard
Normal z:
Step 3. Use the table. The picture shows that we need the cumulative proportion for x =
820. Step 2 says that this is the same as the cumulative proportion for z = −0.99.
The Table A entry for z = −0.99 says that this cumulative proportion is 0.1611. The area
to the right of −0.99 is therefore 1 − 0.1611 = 0.8389.
EXAMPLE 3.8 Who qualifies for college sports?
the National Collegiate Athletic Association uses a sliding scale for eligibility for
Division I athletics. What proportion of all students who take the SAT would meet an
SAT requirement of at least 720, but not 820?
Step 1. State the problem and draw a picture. Call the SAT score x. The variable x has
the N(1026, 209) distribution. What proportion of SAT scores fall between 720 and 820?
Here is the picture:
Step 2. Standardize. Subtract the mean and then divide by the standard deviation to turn
x into a standard Normal z:
Step 3. Use the table. Follow the picture (we added the z–scores to the picture to help
you):
pnorm(-0.99)-pnorm(-1.46) # = 0.0889
About 9% of high school seniors have SAT scores between 720 and 820.
EXAMPLE 3.9 Find the top 10% using software
Scores on the SAT Reading test in recent years follow approximately the N(504, 111)
distribution. How high must a student score to place in the top 10% of all students taking
the SAT?
We want to find the SAT score x with area 0.1 to its right under the Normal curve with
mean μ = 504 and standard deviation σ = 111. That’s the same as finding the SAT score x
with area 0.9 to its left. Figure 3.12 poses the question in graphical form. Most software
will tell you x when you plug in mean 504, standard deviation 111, and cumulative
proportion 0.9. Here is Minitab’s output:
FIGURE 3.12
Locating the point on a Normal curve with area 0.10 to its right, for Examples 3.9 and
3.10.
qnorm(0.9) # = 1.281552
(qnorm(0.9)*111)+504 # = 646.2522
Minitab gives x = 646.252. So scores above 647 are in the top 10%. (Round up because
SAT scores can only be whole numbers.)
EXAMPLE 3.10 Find the top 10% using Table A
Scores on the SAT Reading test in recent years follow approximately the N(504, 111)
distribution. How high must a student score to place in the top 10% of all students taking
the SAT?
Step 1. State the problem and draw a picture. This step is exactly as in Example 3.9.
The picture is Figure 3.12. The x-value that puts a student in the top 10% is the same as
the x-value for which 90% of the area is to the left of x.
Step 2. Use the table. Look in the body of Table A for the entry closest to 0.9. It is
0.8997. This is the entry corresponding to z = 1.28. So z = 1.28 is the standardized value
with area 0.9 to its left.
Step 3. Unstandardize to transform z back to the original x scale. We know that the
standardized value of the unknown x is z = 1.28. This means that x itself lies 1.28
standard deviations above the mean on this particular Normal curve. That is,
A student must score at least 647 to place in the highest 10%.
EXAMPLE 3.11 Find the first quartile
High levels of cholesterol in the blood increase the risk of heart disease. For 14-yearold boys, the distribution of blood cholesterol is approximately Normal with mean μ
= 170 milligrams of cholesterol per deciliter of blood (mg/dl) and standard deviation
σ = 30 mg/dl.8 What is the first quartile of the distribution of blood cholesterol?
Step 1. State the problem and draw a picture. Call the cholesterol level x. The
variable x has the N(170, 30) distribution. The first quartile is the value with 25% of
the distribution to its left. Figure 3.13 is the picture.
Step 2. Use the table. Look in the body of Table A for the entry closest to 0.25. It
is 0.2514. This is the entry corresponding to z = −0.67. So z = −0.67 is the
standardized value with area 0.25 to its left.
Step 3. Unstandardize. The cholesterol level corresponding to z = −0.67 lies 0.67
standard deviations below the mean, so
The first quartile of blood cholesterol levels in 14-year-old boys is about 150 mg/dl.
qnorm(0.25,170,30) # = 149.7653