Transcript Chapter 6: Standard scores and the Normal curve

Chapter 6: Standard Scores and the Normal Curve
Normal distributions show up all over the place.
The standard normal distribution
y
-4
-3
-2
-1
0
z score
1
2
e
 x2

 2




2
3
The standard normal distribution is a continuous distribution.
It has a mean of 0 and a standard deviation of 1
The total area under the curve is equal to 1
4
If we assume that our population is distributed normally, then all we need is the
population mean (m) and standard deviation (s) to convert to z-scores for the standard
normal distribution. We can then find areas under the curve using Table A.
40 45 50 55 60 65 70 75 80 85 90 95 100
Preferred outdoor temperature
Population (X)
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0
z-score
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Standard normal (z)
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All normal distributions can be converted to a ‘standard normal’ by converting to z
scores. We can then look up values in a table in the back of the book (or use a
computer).
Example: Suppose the population of heights of U.S. females is normally
distributed with a mean of 64 inches with a standard deviation of 2.5 inches.
What percent of women are between 61.5 and 66.5 inches?
Answer: 61.5 inches is exactly one standard deviation below the mean, and
66.5 inches is exactly one standard deviation greater than the mean.
Relative frequency
We learned in Chapter 5 that 68% of scores fall within +/- one standard
deviation of the mean, so the answer is 68%
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68%
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-2
-1
0
1
z score
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Using Table A (page 436) to find the area under the standard normal curve
For any given z score, the table gives you two numbers:
Column 2: area between 0 and z
z score
Column 3: area above z
z score
Note that the numbers in columns 2 and 3 always add up to 0.5
Example: what is the proportion of area under the standard normal above z=1?
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-2
-1
0 1
z score
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3
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Look up z=1 in table A (page 436), Column 3. The area is .1587, so 15.87% of the
area of the curve lies beyond one standard deviations above the mean.
Example: Suppose the population of heights of U.S. females is normally
distributed with a mean of 64 inches with a standard deviation of 2.5 inches.
What percent of women are taller than 66.5 inches?
Answer: 65.5 inches is 2.5 inches greater than the mean, which is exactly one
standard deviation. From the last example, we know that the area beyond z=1
is .1587. 15.87% of women are taller than 65.5 inches.
15.87%
54 56.5 59 61.5 64 66.5 69 71.5 74
Height
15.87%
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-1
0
1
z score
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In general, we can calculate the z-score for 60 inches with our formula:
Example: Suppose the mean height of a U.S. female is 64 inches with a
standard deviation of 2.5 inches. What proportion of women are shorter
than 5 feet (60 inches)?
Answer: Convert the height to a z score:
X  mx
60  64
 1.6
sX
2.5
The area below z=-1.6 is equal to the area above z=1.6. Table A, Column 3:
area = .0548. 5.48% of women are shorter than 5 feet.
z

5.48%
5.48%
54 56.5 59 61.5 64 66.5 69 71.5 74
Height
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5.48%
-1
0
1
z score
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-1
0
1
z score
2
3
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Example: Most IQ tests are standardized to have a mean of 100 and a standard
deviation of 15. If we assume that IQ scores are normally distributed, what
percent of the population has ‘very superior intelligence’, or an IQ between 120
and 140?
Over 140 - Genius or near genius
120 - 140 - Very superior intelligence
110 - 119 - Superior intelligence
90 - 109 - Normal or average intelligence
80 - 89 - Dullness
70 - 79 - Borderline deficiency
Under 70 - Definite feeble-mindedness
40
55
70
85
100 115 130 145 160
IQ
Answer: The area between 120 and 140 is equal to the area above 120 minus the
area above 140. z scores for IQs of 120 and 140 are:
z1 
X  mx
sX

120  100
 1.33
15
z2 
X  mx
sX

140  100
 2.67
15
Looking at table A, column 3:
the proportion above z1=1.33 is .0918
the proportion above z2=2.67 is .0038
The difference is .0918-.0038 = .0880.
8.8% of the population has ‘very superior intelligence’
Example: Again, assume IQ scores are normally distributed with m = 100, and s = 15.
For what IQ does 90% of the population exceed?
Answer: We need to use table A backwards.
First we’ll find the z score for which the area beyond z is 0.1. This is a value of z=1.28.
This means that the area below z=-1.28 is 0.1, and the area above z=-1.28 is 0.90.
Next we’ll convert this z score back to IQ scores. This can be done by reversing the
formula that converts from X to z. Solving for X:
Using the new formula: X  u x  zs x  100  (1.28)(15)  80.8
90% of the population has an IQ above 80.8 points
More examples:
The heights of fathers of the 100 students in this class has a mean of 68 inches and a
standard deviation of 3.6 inches. If we assume that this sample is normally distributed:
1) Estimate how many fathers are taller than six feet tall (72) inches?
Answer: Find the z value for a height of 72 inches and use column 3
z
XX
sX

72  68
 1.11
3.6
area =0.1335
Table A, column 3: the proportion
above z=1.11 is .1335
72
# fathers: (100)(.1335)= 13.35
Round to the nearest father : 13
57.2
60.8
64.4
68
z
71.6
75.2
78.8
More examples:
The heights of fathers of the 100 students in this class has a mean of 68 inches and a
standard deviation of 3.6 inches. If we assume that this sample is normally distributed:
2) Estimate how many fathers are shorter than 73 inches.
Answer: Find the z value for a height of 73 inches and use column 2 and add 50%
z
XX
sX

73  68
 1.39
3.6
Table A, Column 2: the proportion
between the mean and z=1.39 is
.4177.
area = .4177
To find the area below z = 1.38, we
need to add another 0.5 to .4177
which gives .9177
#fathers: (100)(.9177)= 91.77
Rounds to 92 fathers
More examples:
The preferred outdoor temperature for the 102 students in this class has a mean of 72.9
degrees and a standard deviation of 9.94 degrees. If we assume that this sample is
normally distributed:
1 How many students prefer a temperature between 55 and 65 degrees?
Answer: Find the z values for 55 and 65 and use table A, column 3
z1 
z2 
X 1  mx
sX
X 2  mx
sX


55  72.9
 1.8
9.94
area = .1789
65  72.9
 0.79
9.94
To find the area below z1 = -1.8, we find the
area above z = 1.8 which is .0359
The area below z2 = -.9, is .2148
The area between z1 and z2 is .2148- .0359 = .1789
Rounding to the nearest student: (0.1789)(102) = 18 students
More examples:
The preferred outdoor temperature for the 102 students in this class has a mean of 72.9
degrees and a standard deviation of 9.94 degrees. If we assume that this sample is
normally distributed.
2) For what temperature do 90% of the preferred temperatures fall below?
Answer: Find the z value that corresponds to the top 90% of the standard normal Then
find the corresponding temperature using:
X  u x  zs x
Using table A backwards, the z-score for an
area of 0.9 is z = 1.28.
Converting to temperature, 72.9+(1.28)(9.94) =
85.6 degrees.