Normal Distribution

1. NORMAL DISTRIBUTION
68% – 95% – 99.7% Rule

2. 68-95-99.7 Rule
68% of
the data
95% of the data
99.7% of the data

3. Properties of Normal Distributions
Normal distribution
• A continuous probability distribution for a random
variable, x.
• The most important continuous probability
distribution in statistics.
• The graph of a normal distribution is called the
normal curve.
x
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Larson/Farber 4th ed

4. Properties of Normal Distributions
1. The mean, median, and mode are equal.
2. The normal curve is bell-shaped and symmetric
about the mean.
3. The total area under the curve is equal to one.
4. The normal curve approaches, but never touches
the x-axis as it extends farther and farther away
from the mean.
x
Total area = 1
μ
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Larson/Farber 4th ed

5. The Standard Normal Distribution
Standard normal distribution
• A normal distribution with a mean of 0 and a standard
deviation of 1.
3 1
2 1 0 2 3
z
Area = 1
Value - Mean -
Standard deviation
x
z


 
• Any x-value can be transformed into a z-score by
using the formula
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Larson/Farber 4th ed

6. The Standard Normal Distribution
• If each data value of a normally distributed random
variable x is transformed into a z-score, the result will
be the standard normal distribution.
Normal Distribution
x
m
s
m=0
s=1
z
Standard Normal
Distribution
-
x
z



• Use the Standard Normal Table to find the
cumulative area under the standard normal curve.
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Larson/Farber 4th ed

7. Properties of the Standard Normal
Distribution
1. The cumulative area is close to 0 for z-scores close
to z = 3.49.
2. The cumulative area increases as the z-scores
increase.
z = 3.49
Area is
close to 0 z
3 1
2 1 0 2 3
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Larson/Farber 4th ed

8. z = 3.49
Area is close to 1
Properties of the Standard Normal
Distribution
3. The cumulative area for z = 0 is 0.5000.
4. The cumulative area is close to 1 for z-scores close
to z = 3.49.
Area is 0.5000
z = 0
z
3 1
2 1 0 2 3
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Larson/Farber 4th ed

9. Example
Suppose JAM scores roughly follows a normal
distribution in the Indian population of college-bound
students (with range restricted to 200-800), and the
average JAM Score be 500 with a standard deviation
of 50, then:
68% of students will have scores between
450 and 550
95% will be between
400 and 600
99.7% will be between
350 and 650

10. 68-95-99.7 Rule
68% of
the data
95% of the data
99.7% of the data

11. The Standard Normal Distribution (Z)
All normal distributions can be converted into the
standard normal curve by subtracting the mean and
dividing by the standard deviation:




X
Z
Somebody calculated all the integrals for the standard normal
and put them in a table. So we never have to integrate!
Even better, computers now do all the integration.

12. The Standard Normal (Z):
“Universal Currency”
The formula for the standardized normal
probability density function is
2
2
)
(
2
1
)
1
0
(
2
1
2
1
2
)
1
(
1
)
(
Z
Z
e
e
Z
p










13. Comparing X and Z units
Z
100
2.0
0
200 X ( = 100,  = 50)
( = 0,  = 1)

14. Example
• For example: What’s the probability of getting a math JAM score of 575 or less,  = 500 and 
= 50?
5
.
1
50
500
575



Z
i.e., A score of 575 is 1.5 standard deviations above the mean

 











5
.
1
2
1
575
200
)
50
500
(
2
1 2
2
2
1
2
)
50
(
1
)
575
( dz
e
dx
e
X
P
Z
x


But to look up Z = 1.5 in standard normal table = 0.9332

15. Practice problem
If birth weights in a population are normally distributed with a
mean of 109 oz and a standard deviation of 13 oz,
a. What is the chance of obtaining a birth weight of 141 oz or
heavier when sampling birth records at random?
46
.
2
13
109
141



Z
From the chart  Z of 2.46 corresponds to a right tail (greater than) area
of: P(Z≥2.46) = 1-P(Z≤2.46) = 1-(0.9931)= 0.0069 or 0.69 %

16. Practice problem
b. What is the chance of obtaining a birth weight of 120 or lighter?
From the chart  Z of 0.85 corresponds to a left tail area of:
P(Z≤0.85) = 0.8023= 80.23%

17. Looking up probabilities in the standard
normal table
What is the area to
the left of Z=1.51 in a
standard normal
curve?
Z=1.51
Z=1.51
Area is
93.45%

18. Finding Areas Under the Standard Normal
Curve
1. Sketch the standard normal curve and shade the
appropriate area under the curve.
2. Find the area by following the directions for each
case shown.
a. To find the area to the left of z, find the area that
corresponds to z in the Standard Normal Table.
1. Use the table to find
the area for the z-score
2. The area to the left
of z = 1.23 is
0.8907
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Larson/Farber 4th ed

19. Finding Areas Under the Standard Normal
Curve
b. To find the area to the right of z, use the Standard
Normal Table to find the area that corresponds to z.
Then subtract the area from 1.
3. Subtract to find the area
to the right of z = 1.23:
1  0.8907 =
0.1093.
1. Use the table to find the
area for the z-score.
2. The area to the
left of z = 1.23
is 0.8907.
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Larson/Farber 4th ed

20. Finding Areas Under the Standard Normal
Curve
c. To find the area between two z-scores, find the area
corresponding to each z-score in the Standard Normal
Table. Then subtract the smaller area from the larger
area.
4. Subtract to find the area of
the region between the two
z-scores:
0.8907  0.2266 = 0.6641.
3. The area to the
left of z = 0.75 is
0.2266.
2. The area to the
left of z = 1.23
is 0.8907.
1. Use the table to find the
area for the z-scores.
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Larson/Farber 4th ed

21. Example: Finding Area Under the Standard
Normal Curve
Find the area under the standard normal curve to the
left of z = -0.99.
From the Standard Normal Table, the area is
equal to 0.1611.
0.99 0
z
0.1611
Solution:
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Larson/Farber 4th ed

22. Example: Finding Area Under the Standard
Normal Curve
Find the area under the standard normal curve to the
right of z = 1.06.
From the Standard Normal Table, the area is equal to
0.1446.
1  0.8554 = 0.1446
1.06
0
z
Solution:
0.8554
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Larson/Farber 4th ed

23. Find the area under the standard normal curve
between z = 1.5 and z = 1.25.
Example: Finding Area Under the Standard
Normal Curve
From the Standard Normal Table, the area is equal to
0.8276.
1.25
0
z
1.50
0.8944
0.0668
Solution:
0.8944 0.0668 = 0.8276
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Larson/Farber 4th ed