Chapter 6 Page 1 of 14
Chapter 6
Normal Curve
Spring 2016
(12/26/15)
Chapter 6 Page 2 of 14
Chapter 17 Normal Distribution
Draw a histogram bar graph of height of 501 basketball players from 2012 frequency
table player rosters.
height 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87
in.
number 3 12 20 24 38 33 34 41 45 50 56 52 41 33 11 2 1
players
60
50
40
30
20
10
0
71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87
Height
If it has the general appearance resembling a bell-shape, it is called a normal distribution or
normal curve, and there will be extra information that can be interpreted from the graph for
predicting the height of basketball players on future teams.
From previous chapter, if we calculate µ (median),
µ
σ
σ (standard deviation)
= 79
= 3.47
you can make predictions about height of future year players.
Chapter 6 Page 3 of 14
symmetry, µ (median),
σ (standard deviation) for basketball players
symmetrical
symmetrical
Curve change point P
located 1 σ left of median
Curve change point P
located 1 σ right of median
σ = 3.47 σ = 3.47
0
75.53”
79”
82.47”
µ = median
Height
50% of future players will be 79” or taller
(and 50% will be 79” or shorter)
50%
79” height
Quartiles: (25%, 75%)
µ - .675 σ = 79 - .675(3.47) = 76.66”
Q3 = µ + .675 σ = 79 + .675(3.47) = 81.34”
Q1 =
25% future players
will be 76.66” or shorter
75% future players
will be 81.34”
or shorter
25% will be
between 76.66”
and 79”
50% future players will be between Q1 and Q3 tall
(76.66” and 81.34”)
25% future players will be 81.34” or taller
25%
75% future players will be 76.66” or taller
25%
50%
25%
25%
76.66 79 81.34 height
Q1
Q3
25% will be between 79” and
81.34”
Chapter 6 Page 4 of 14
Standardizing Normal Data - means describing player heights on the horizontal axis
in terms of “z” (distance in standard deviations 3.47”)
1 “z” = σ = 3.47 inches
curvature change point P
σ
3.47
curvature change point P
σ
3.47
σ
3.47
68.59
72.06
75.53
z = -3
z = -2
z = -1
79
μ
z=0
σ
3.47
σ
3.47
82.47
z = +1
σ
3.47
85.94
89.41” tall
z = +2
z = +3
79” + 3.47” = 82.47” is (1) σ to the right of the median or z = +1
79” + 2(3.47”) = 85.94” is (2) σ to the right of the median or z = +2
79” + 3(3.47”) = 89.41” is (3) σ to the right of the median or z = +3
79” - 3.47” = 75.53” is (1) σ to the left of the median or z = -1
79” - 2(3.47”) = 72.06” is (2) σ to the left of the median or z = -2
79” - 3(3.47”) = 68.59” is (3) σ to the left of the median or z = -3
What is 6 foot (72”) in z’s ? 72 – 79 = -7
-7/3.47 = -2.02
z = -2.02
What is 7 foot (84”) in z’s ? 84 – 79 = +5
+5/3.47 = +1.44
z = +1.44
Chapter 6 Page 5 of 14
Practice problems page 193
3.) Given following graph: with 32 and 48 located at the curvature change points….
Curve change points P
32
Find:
48
µ, σ, Q3, Q1
a.) The median µ is located midway in the distance between the curvature change
points 32 and 48………32 + 48 = 40 (½ way point)
2
50%
50%
P
P
8

32
8

40
48
b.) The standard deviation σ is the distance from the median to the curvature change point
µ = 40; curvature change point is at 48; σ is 48 – 40 = 8
c.) Q3 =  + .675  = 40 + .675(8) = 45.4
¼ points, 25% each
d.) Q1 =  - .675  = 40 - .675(8) = 34.6
25%
25%
25%
25%
8

8

32
40
48
34.6 µ 45.4
Q1
Q3
Chapter 6 Page 6 of 14
17.) Given following graph: with µ = 310 and Q3 = 391
310
µ
a.) What is “z” for 490?
First, find σ: Q3 =
µ + .675 σ
391 = 310 + .675 σ
391 – 310 = σ = 120
.675
“z” for 490: 490 – 310 = + 1.5 (means 490 is 1.5 σ
120
b.) What is “z” for 250?
to the right of
µ)
“z” for 250: 250 – 310 = - 0.5
120
z = - 0.5 (means 250 is .5 σ to the left of
c.) What is “z” for 220?
µ)
“z” for 220: 220 – 310 = - 0.75
120
z = - 0.75 (means 220 is .75 σ to the left of
d.) What is “z” for 442?
µ)
“z” for 442: 442 – 310 = + 1.1
120
z = +1.1 (means 442 is 1.1 σ to the right of
µ)
Chapter 6 Page 7 of 14
(68-95-99.7) RULE - property of normal curves. All Normal curves have the same geometrical
characteristics defined by (68-95-99.7) RULE
Example: height of 501 basketball players from 2012 frequency table player rosters.
µ = 79”
σ = 3.47”
68 95 99.7 RULE
99.7 % = 2.35%+13.5%+34%+34%+13.5%+2.35%
95% = 13.5%+34%+34%+13.5%
68% = 34%+34%
Curve change point
1 σ left of median
0.15% 2.35%
players
Curve change point
1 σ right of median
34% of
players
 = 3.47
34% of
players
 = 3.47
9% 25%
25% 9%
13.5%
players
( - 2)
72.06
2.5th percentile
16th percentile
25th percentile
50th percentile
75th percentile
84th percentile
97.5th percentile
99.85th percentile
0.15%
13.5%
players
Q1
( -3 )
68.59
2.35%
players
( - )
75.53
76.66”
Q3
Median  ( + )
79
82.47
( + 2)
85.94
( + 3)
89.41” tall
81.34”
important percentiles
add %’s starting from far left; example: 0.15% + 2.35% + 13.5% =
16th percentile
Chapter 6 Page 8 of 14
Example- BRICKLAYERS - the American Bricklayers Association, who compiles data for
thousands of bricklayers, published the following data for Contracting firms to use for estimating
labor requirements:
“The median bricklayer can install 60 bricks per hour,
and the standard deviation for bricklayers is 5 bricks per hour.”
μ = 60
σ= 5
Pretend you work for ABC Construction Co. who will hire 1000 bricklayers to build
several new dorms and other buildings on New York campuses.
The Project Superintendent asks you the following typical construction questions:
1.) How many bricklayers can lay more than 65 bricks in one hour? Pay them
“premium wages” for advanced workers.
2.) How many can lay between 45 and 70 bricks every hour? Assign those men to
DORM B.
3.) How many bricklayers will lay less than 45 bricks in one hour? Set up a training
program for them.
4.) How many bricklayers will we have for DORM C where we need to lay 70 or
more bricks every hour?
5.) How many bricklayers can lay between 55 and 65 bricks every hour? We’ll pay
them the standard hourly wage rate for Journeymen.
Without knowing anything about Bricks, Bricklaying, or Bricklaying Craftsmen, you can
answer all these questions by knowing Normal Curve 68-95-99.7 characteristics.
Chapter 6 Page 9 of 14
The Bricklayer Association Normal Curve for this data looks like this: 68
95
99.7
99.7 % = 2.35%+13.5%+34%+34%+13.5%+2.35%
95% = 13.5%+34%+34%+13.5%
68% = 34%+34%
Curve change point
Curve change point
34% of
34% of
bricklayers bricklayers


0.15% 2.35%
b’layers
9% 25%
13.5%
b’layers
25% 9%
2.35%
b’layers
0.15%
13.5%
b’layers

Q3
50
55
60
65
70
75
BRICKS / HR.
Q1 =  - .675  (25th percentile)
= 60 - .675(5)
Q3 =  + .675  (75th percentile)
= 57 bricks/hr.
= 60 + .675(5)
= 63 bricks/hr.
Q1
45
1.) How many of the 1000 bricklayers will lay 65 bricks/hour or more
= (13.5% + 2.35% + .15%) of 1000 = 160 bricklayers
2.) How many of the 1000 bricklayers will lay 45 bricks/hour to 70 bricks/hour
= (2.35% + 13.5% + 34% +34% + 13.5%) of 1000 = 974 bricklayers
3.) How many of the 1000 bricklayers will lay 45 bricks/hour or less
= .15% of 1000 = 2 bricklayers
4.) How many of the 1000 bricklayers will lay 70 bricks/hour or more
= (2.35% + .15%) of 1000 = 25 bricklayers
5.) How many of the 1000 bricklayers will lay from 55 bricks/hour to 65 bricks/hour
= (34% + 34%) of 1000 = 680 bricklayers
Do practice problems page 194: 29, 31ab
Chapter 6 Page 10 of 14
Golf – The PGA (Professional Golfers Association) say that at Battle Island Golf Course in
Oswego, the median score is 101, and the standard deviation is 8.
 = 101
=8
One of your new jobs at work is being the Assistant Golf League Secretary for the Thursday
Night League. There are 30 ladies playing this Summer. You don’t know any of them personally.
In fact, you don’t even play golf. Still…you can forecast:
1.) How many of the ladies will score 101 or more?
2.) How many of the ladies will score 93 to 101?
3.) How many of the ladies will score 85 to 96?
4.) How many of the ladies will score 85 or less?
5.) How many of the ladies will score 117 or more?
6.) What is the score of the golfer at the 1st Quartile (25th percentile)?
7.) How many golfers will be in the Inter Quartile Range?
8.) How many golfers will score 93 or less?
for
prizes
Chapter 6 Page 11 of 14
The Golf Normal Curve looks like this:
68
95
99.7
99.7 % = 2.35%+13.5%+34%+34%+13.5%+2.35%
95% = 13.5%+34%+34%+13.5%
68% = 34%+34%
Inflection change point
Inflection change point
34% of
golfers

0.15% 2.35%
golfers
9%
13.5%
golfers
25%
34% of
golfers

25% 9%
2.35%
golfers
0.15%
13.5%
golfers

Q3
77
85
93
101 106 109
117
125
Golf Score
Q1 =  - .675  (25th percentile)
= 101 - .675(8)
Q3 =  + .675  (75th percentile)
= 96 strokes .
= 101 + .675(8)
= 106 strokes
1.) How many of the ladies will score 101 or more?
= (34% + 13.5% + 2.35% + .15%) of 30 = 50% of 30 = 15
2.) How many of the ladies will score 93 to 101? shaded
= (34%) of 30 = 10.2 = approximately 10
3.) How many of the ladies will score 85 to 96?
= (13.5% + 9%) of 30 = 22.5% of 30 = 6.75 = approximately 7
4.) How many of the ladies will score 85 or less? shaded
= (0.15% + 2.35%) of 30 = 2.5% of 30 = .75 = approximately 1
5.) How many of the ladies will score 117 or more? shaded
= (2.35% + 0.15%) of 30 = 2.5% of 30 = .75 = approximately 1
6.) What is the score of the golfer at the 1st Quartile (25th percentile, Q1)?
Q1 =  - .675  = 101 - .675(8) = 96
7.) How many golfers will be in the Inter Quartile Range?
= (25% + 25%) of 30 = 50% of 30 = 15
8.) How many the ladies will score 93 or less?
= (13.5% + 2.35% + 0.15%) of 30 = 16% of 30 = 4.8 = approximately 5
Q1
96
Do practice problems page 194: 41abcd,42abc,43ab,44abcd
Chapter 6 Page 12 of 14
Practice problems page 194
29.) Given: 68%, between 42, 54. Find  and  for following graph:
All Normal curves have the same geometrical characteristics defined by (68-95-99.7) RULE
68% = 34%+34%
Curve change point
Curve change point
located above 1 σ right of median
34% of
data

34% of
data

( - )
42
Median 
( + )
Mean
54
2 σ = 54 - 42 = 12
σ = 12/2 = 6
µ = 42 + 6 = 48
31a,b.) Given: 95%, between 85.2, 98.8. Find   Q1 Q3 for following graph:
All Normal curves have the same geometrical characteristics defined by (68 - 95 - 99.7) RULE
95% = 13.5%+34%+34%+13.5%
Curve change point
Curve change point
13.5%

( - 2)
85.2
34% of
data

( - )
4 σ = 98.8 - 85.2 = 13.6
σ = 13.6/4 = 3.4
µ = 85.2 + 3.4 + 3.4 = 92
34% of
data

Median 
Mean
13.5%

( + )
( + 2)
98.8
Q1 =  - .675  = 92 - .675(3.4) = 89.71
Q3 =  + .675  = 92 + .675(3.4) = 94.30
Chapter 6 Page 13 of 14
Practice problems page 195
Given:µ
= 52
 = 11
41.)
2500
Students
68 - 95 - 99.7
Inflection change point
Inflection change point
34%
students
0.15% 2.35%
students
13.5%
students
30
41
19
2.5th
16th percentile
25th percentile
50th percentile
75th percentile
84th percentile
97.5th percentile
99.85th percentile
9%
34%
students
25%
25% 9%
2.35%
students
0.15%
13.5%
students
Q1
(45)
52
Q3 63
(59)
74
85
41a.) Average Score = Mean = Median = 52
41b.) Score 52 or more = 34% + 13.5% + 2.35% + .15% = 50% of students
(.5 X 2500 = 1250 students)
41c.) Score 41 - 63 = 34% +34% = 68% of students
(.68 X 2500 = 1700 students) shown shaded
41d.) Score 63 or more = 13.5% + 2.35% + .15% = 16% of students
(.16 X 2500 = 400 students)
42a.) Score 30 - 74 = 13.5% + 34% + 34% + 13.5% = 95% of students
(.95 X 2500 = 2375 students)
42b.) Score 30 or less = .15% + 2.35% = 2.5% of students
(.025 X 2500 = 63 students) shown shaded
Chapter 6 Page 14 of 14
42c.) Score 19 or less = .15% of 2500students
(.0015 X 2500 = 4 students)
43a.) Score Q1 = µ - .675σ = 52 - .675(11) = 45 points
43b.) Score Q3 = µ + .675σ = 52 + .675(11) = 59 points
(IQR = Q3 - Q1 = 59 - 45 = 14 points)
44a.) Score 52 = 50th percentile
44b.) Score 63 = 84th percentile
44c.) skip this question
44d.) Score 85 = 99.85th percentile
this questions wasn’t asked
but I included it for
practice