Some material used was contributed by Tom Koesters – East Forsyth High School
CHAPTER 16 – NORMAL CURVES
No quartile statements
Basic:
Identify Mean/Median, Standard Deviation
z-value (z-score) calculation back and forth
68, 95, 99.7 Rule
Some material used was contributed by Tom Koesters – East Forsyth High School
Section 16.2 – Normal Distributions
Normal Curves and Normal Distributions
A set of data whose distribution has a bell-shaped curve is said to have a NORMAL DISTRIBUTION.
A perfect bell-shaped curve is called a NORMAL CURVE.
The area under the curve is equal to 1.0 probability or 100%.
 SHAPE: Symmetry: A normal curve is symmetrical along a vertical line splitting the bell in half.
 CENTER: Mean equals Median:
In a perfectly normal distribution, the line of symmetry
is the median M and the mean μ of the data.
o 50% of the data are greater than or equal to the mean
o 50% are less than or equal to the mean.
P
P
σ
μ–σ
μ+σ
 Standard Deviation and Point of Inflection. A point of inflection P on a curve is the point
where the concavity of the curve changes, and is one standard deviation σ away from the center (mean
or median) of the data. P = μ ± σ
Example #1: Find the mean, median and standard deviation of each normal distribution shown. The
point of inflection of the curve is P is the point of inflection of the curve.
a.
μ = ___78_____
b.
μ = ___82_____
c.
μ = __87______
M = __78_____
M = __82_____
M = __87_____
σ = _96 – 78 = 18
σ = _99 – 82 = 17
σ = _93 – 87 = 16
Example #2: Consider the following normal distributions where P and P′ are the points of inflection of
the curve. Find the mean and standard deviation. (round to the nearest tenth).
2a.
μ = _70 = (76+64)/2_
σ = __6______
2b.
μ = _79 = (86+72)/2_
σ = __7_______
2c.
μ = _79 = (87+71)/2_
σ = __8_______
Some material used was contributed by Tom Koesters – East Forsyth High School
Section 16.3 – Standardizing Normal Data
Standardized Value, Z-Value, or Z-Score:
To standardize the comparison of data in we refer to data values based on the number of standard deviations
they are above or below the mean. This standardized value is often called the z-value or z-score.
Symbol Example:
If a data value Xi is exactly one standard deviation above the mean, its z-value is z = 1.0. If a data value Xi is
exactly one standard deviation below the mean, its z-value is z = –1.0.
Data Example:
If mean is 8 and standard deviation is 3, then data point of 11 has z = 1.0 and the data value of 5 has z = -1.0.
z-value Calculation: z 
X 

The z-value calculation is the difference of the data value (X) and mean (μ) all over the standard deviation (σ).
Example #1: Consider a normally distributed data set with a mean μ = 45 ft. and a standard deviation σ
= 10 ft. Find the z-value for the given data values.
55  45
50  45
1a) X = 55 ft. z 
1c) X = 50 ft. z 
 1.0
 0.5
10
10
55 ft. is one σ above the mean
50 ft. is one-half σ above the mean
20  45
 2.5
10
35 ft. is two and a half σ below the mean
1b) X = 20 ft. z 
21.58  45
 2.342
10
21.58 ft. is 2.342 σ below the mean
1d) X = 21.58 ft.
z
Example #2: Consider a normally distributed data set with μ = 18.5 and σ = 1.2. Find the z-values.
2a) X = 22.1
22.1  18.5
z
 3.0
1.2
2b) X = 16.34
16.34  18.5
z
 1.8
1.2
3c) X = 19.3
19.3  18.5
z
 0.66
1.2
Solving for data point given the standardized value: z-value to X.
Example #3: Consider a normally distributed data set with mean μ = 235.7 meters and standard
deviation σ = 41.58 meters. Find the data value X that corresponds to the standardized value
3a) z = –3.45?
3b) z = 2.6?
3c) z = 0.3?
Above
Mean
by
2.6
standard
X  235.7
X  235.7
 3.45
 0.3
deviations
41.58
41.58
235.7  2.6(41.58)
X  92.249m
X  248.174m
X  343.808m
Example #4: Consider a normally distributed data set with μ = 12.5 and σ = 0.35. Find data value X that
corresponds to the standardized value?
4a) z = –3.45?
4b) z = 2.6?
4c) z = 0.3?
3.45 standard deviation below
X  12.5
X  12.5
 2.6
 0.3
Mean
.35
.35
12.5  3.45(.35)  11.2925
X  13.41
X  12.605
Some material used was contributed by Tom Koesters – East Forsyth High School
Section 16.4 68-95-99.7 Rule
99.7%
95%
68%
In a set of data with a normal distribution, approximately
 68% of the data fall within one standard deviation of the mean (between z = –1 and z = 1)
 95% of the data fall within two standard deviations of the mean (between z = –2 and z = 2
 99.7% of the data fall within three standard deviations of the mean (between z = –3 and z = 3)
1. Consider the normal distribution shown in the graph.
1a. Find the mean and the standard deviation of the distribution.
μ = __59 = (65+53)/2__
σ = ____6____
1b. What percent of the data is greater than X = 65?
16% = (100 – 68)/2
1c. Suppose the size of the population is N = 114. Approximately how many data values are less than 53?
0.16(114) =18.24 approximately 18 data values
2. Consider the normal distribution shown in the graph.
2a. Find the mean and the standard deviation of the distribution.
μ = __59 = (66+52)/2__
σ = __7______
2b. What percent of the data is greater than X = 52?
84% = 100 – 16% or 68% +_ 16%
2c. Suppose the size of the population is N = 127. Approximately how
many data values are there between 52 and the mean?
.34(127) = 43.18 approximately 43 data values
Some material used was contributed by Tom Koesters – East Forsyth High School
3. Consider the normal distribution shown in the graph.
3a. Find the mean of the distribution.
μ = __76___
3b. What percent of the data is greater than X = 79.75?
25% = 100% - 75%
3c. Suppose the size of the population is N = 89. Approximately how
many data values are there between 72.25 and 79.75?
0.5(89) = 44.5 approximately 45 data values
If normal curve is not provided, use number line for the data since z-value and percentages never change.
Data Values
μ – 3σ
0.15%
Z-values
μ – 2σ
μ– σ
μ
μ+σ
μ + 2σ μ + 3σ
2.35% 0.15%
13.5%
34%
34%
13.5%
z = -2
z = -1
z=0
z=1
z=2 z=3
2.35%
z = -3
5. A normal distribution has a standard deviation σ = 6.1 cm and 84% of the data fall above 47.1 cm.
Find the mean.
μ = _ 53.2 = 47.1 + 6.1_
16% below  X = 47.1  84% above
0.15 + 2.35 + 13.5 = 16%, so 47.1  z = -1 is one standard deviation below mean
6. A normal distribution has a standard deviation σ = 9.6 cm and 84% of the data fall below 49.7 cm.
Find the mean.
μ = _40.1 = 49.7 – 9.6 _
16% below  X = 49.7  84% above so 49.7 is one standard deviation above mean
7. A normal distribution has mean μ = 11.8 and standard deviation σ = 3.4. Approximately what percent
of the data fall between 1.6 and 22.0?
1.6 = 11.8 – 3.4(3) = three standard deviations below mean
22.0 = 11.8 + 3.4(3) = three standard deviations above mean
Between 3 Standard Deviations = 99.7%
8. A normal distribution has mean μ = 12.5 and standard deviation σ = 3.1. Approximately what percent
of the data fall between 6.3 and 18.7?
6.3 = 12.5 – 3.1(2) = two standard deviations below mean
18.7 = 12.5 + 3.1(2) = two standard deviations above mean
Between 2 Standard Deviations = 95%
9. In a normal distribution with mean μ and standard deviation σ, what percent of the data
a. fall below the value μ + σ? __84%_ b. between μ – 3σ and μ – 2σ? _2.35%_ c. above μ – 3σ? _99.85%
Some material used was contributed by Tom Koesters – East Forsyth High School
Chapter 16 – Normal Curve Practice Problems
1. Find the z-value for a given data point X in a normal distribution of data with mean μ = 100 and
standard deviation σ = 10 (round your answers to the nearest tenth).
1a.
X = 140
z = __4______
1b.
X = 112
z = __1. 2_____
1c.
X = 139
z = __3.9_____
1d.
X = 83
z = __-1.7_____
2. Find the data point X that corresponds to the given z-value in a normal distribution of data with μ =
180.3 ft. and standard deviation σ = 30.9 ft. (round your answers to the nearest tenth).
2a.
z = –4
X = ___56.7__
2b.
z = 0.8
X = ___205.02___
2c.
z = 4.5
X = ___319.35_
2d.
z=0
X = ___180.3____
3. In a normal distribution, the data value X1 = 30 has the standardized value z1 = –2 and the data the
data value X2 = 130 has the standardized value z2 = 3. Find the mean and standard deviation.
130  30 = 5 standard deviations … σ = 20 and μ = 70
4. In a normal distribution, the data value X1 = 30 has the standardized value z1 = –2 and the data the
data value X2 = 75 has the standardized value z2 = 1. Find the mean and standard deviation.
75  30 = 3 standard deviations … σ = 15 and μ = 60
5. In a normal distribution, the data value X1 = 20 has the standardized value z1 = –3 and the data the
data value X2 = 115 has the standardized value z2 = 2. Find the mean and standard deviation.
115 20 = 5 standard deviations … σ = 19 and μ = 77
6. Consider the normal distribution shown in the graph.
6a. Find the mean of the distribution.
μ = ___84____
6b. What percent of the data is less than X = 95.50? ___75%__
7) The number of problems missed on a quiz follows a normal distribution with a mean of 15 and a
standard deviation of 4.
7a. What percent of students
7b. What percent of students
7c. What percent of students less
missed between 11 – 19
missed more than 23 problems?
than 23 problems?
problems? 68%
2.5%
97.5%
8) The final exam scores normal distribution with a mean of 80 and a standard deviation of 6.
8a. What percent of students
8b. What percent of students
8c. What percent of students
scored between 80 - 92?
scored between 68 -86?
scored higher than 74?
47.5%
81.5%
84%
9) The class mean of height is 60 inches. We know that 68% of the students are between 56 and 64 inches.
9a. What is the standard deviation?
9c. 16% of students are taller than …?
4 in
64 in
9b. 50% of the students are shorter than …?
60 in
9d. 84% of students are taller than …?
56 inches
Honors Discrete Practice Multiple Choice
CHAPTER 16: NORMAL CURVE
Pick the MOST ACCURATE Answer Choice
For #1 – 11: 250 students in a math class take the final exam. The scores on the exam have
an approximately normal distribution with center μ = 75 and standard deviation σ = 10.
1) The number of students scoring 75 or more is approximately
A. 75
B. 83
C. 125
D. 158
2) The average score on the exam was approximately
A. 10
B. 75
C. 85
D. 95
3) Approximately 95% of the class scored between
A. 55 and 95
B. 0 and 95
C. 45 and 100
D. 65 and 85
4) Assuming there were no outliers, the lowest score on the exam was around
A. 75
B. 83
C. 125
D. 158
5) Approximately what percent of the students scored between 65 and 75?
A. 10%
B. 34%
C. 68%
D. 95%
6) Peter’s score on the exam places him in the 16th percentile of the class. Peter’s score
on the exam is approximately?
A. 16
B. 45
C. 55
D. 65
7) Carol scored 85 points on the exam. In approximately what percentile does this
score place her?
A. 34th
B. 68th
C. 84th
D. 95th
8) A z-value of 1.8 corresponds to a test score of
A. 18
B. 57
C. 76.8
D. 93
9) A score of 95 corresponds to a standardized value (z-value) of
A. – 2
B. 2
C. 8.5
D. 20
10) A score of 60 corresponds to a standardized value (z-value) of
A. – 5
B. – 1.5
C. 1.5
D. 5
11) Approximately what percentage of test scores had standardized values between -3
and 3?
A. 50%
B. 68%
C. 95%
D. 99%
For #12 – 14: 95% of the data for a normal curve is between data values of 24 and 84.
12) Find the mean, μ, of the normal curve
A. 12
B. 39
C. 48
D. 54
13) Find the standard deviation, σ, of the normal curve
A. 10
B. 15
C. 30
D. 60
14) What z-value represents the data value 84?
A. 2.0
B. 3.0
C. 15
D. 30
For #15 – 17: 16% of the data for a normal curve is above the data value 56 and another
16% of the data for the normal curve is below the data value of 44.
15) Find the mean, μ, of the normal curve
A. 38
B. 48
C. 50
D. 62
16) Find the standard deviation, σ, of the normal curve
A. 3
B. 6
C. 12
D. 24
17) What z-value represents the data value 44?
A. – 1.0
B. 1.0
C. – 2.0
D. 2.0
For # 18 - 21: As part of a study, 800 college football players are randomly chosen and
their weights taken. The distribution of weights is approximately normal. The average
weight is 235 pounds and the standard deviation is 25 pounds.
18) Approximately how many players weighed 210 pounds or less?
A. 64
B. 128
C. 200
D. 544
19) Assuming there were no outliers, the range of player weights was approximately
A. 100 lbs
B. 150 lbs
C. 200 lbs
D. 400 lbs
20) Approximately what percentage of players weighed over 285 pounds?
A. 95%
B. 97.5%
C. 2.5%
D. 5%
21) Approximately how many players weighed between 210 and 260 pounds?
A. 760
B. 128
C. 256
D. 544
Honors Discrete Practice Multiple Choice
CHAPTER 16: NORMAL CURVE – Solutions
1) C
2) B
3) A
4) C
5) B
6) D
7) C
8) D
9) B
10) B
11) D
12) D
13) B
14) A
15) C
16) B
17) A
18) B
19) B
20) C
21) D