2. 1-1
The Normal
Distribution
GOALS
ONE
List the characteristics of the normal distribution.
TWO
Define and calculate z values.
THREE
Determine the probability that an observation will lie between two points
using the standard normal distribution.
FOUR
Determine the probability that an observation will be above or below a given
value using the standard normal distribution.
Jonathan M. Pacurib
3. 1-1
GOALS
FIVE
Compare two or more observations that are on different probability
distributions.
Jonathan M. Pacurib
4. 7-3
Characteristics of a Normal
Distribution
The normal curve is bell-shaped and has a single
peak at the exact center of the distribution.
The arithmetic mean, median, and mode of the
distribution are equal and located at the peak.
Half the area under the curve is above the peak,
and the other half is below it.
5. 7-4
Characteristics of a Normal
Distribution
The normal distribution is symmetrical
about its mean.
The normal distribution is asymptotic - the
curve gets closer and closer to the x-axis
but never actually touches it.
6. 7-5 r a l i t r b u i o n : µ = 0 , σ2 = 1
Characteristics of a Normal Distribution
0 . 4
Normal
curve is
0 . 3 symmetrical
0 . 2
Theoretically,
curve
f ( x
0 . 1
extends to
infinity
. 0
- 5
a
Mean, median, and
x
mode are equal
7. Properties of a Normal Distribution
Inflection point Inflection point
x
• As the curve extends farther and farther away from the
mean, it gets closer and closer to the x-axis but never
touches it.
• The points at which the curvature changes are called
inflection points. The graph curves downward between the
inflection points and curves upward past the inflection
points to the left and to the right.
8. Means and Standard Deviations
Curves with different means, same standard deviation
10 11 12 13 14 15 16 17 18 19 20
Curves with different means, different standard deviations
9 10 11 12 13 14 15 16 17 18 19 20 21 22
10. 7-6
The Standard Normal
Distribution
A normal distribution with a mean of 0 and a
standard deviation of 1 is called the standard
normal distribution.
Z value: The distance between a selected
value, designated X, and the population mean
, divided by the population standard
µ
deviation,
σ
X −µ
Z =
σ
11. The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1 0 1 2 3 4 z
12. The Standard Score
The standard score, or z-score, represents the number of
standard deviations a random variable x falls from the
mean.
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of
7. Find the standard z-score for a person with a score of:
(a) 161 (b) 148 (c) 152
(a) (b) (c)
13. From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of 7.
Find the test score for a person with a standard score of:
(a) 2.33 (b) –1.75 (c) 0
(a) x = 152 + (2.33)(7) = 168.31
(b) x = 152 + (–1.75)(7) = 139.75
(c) x = 152 + (0)(7) = 152
14. Cumulative Areas
The
total
area
under
the curve
is one.
–3 –2 –1 0 1 2 3 z
• The cumulative area is close to 0 for z-scores close
to –3.49.
• The cumulative area for z = 0 is 0.5000.
• The cumulative area is close to 1 for z-scores close to
3.49.
15. References:
Bluman. Allan G. Elementary Statistics.
Step by Step. McGraw Hill international
Edition, 2008.
Pacho, Elsie et. Al. Fundamental Statistics.
Trinitas Publishing House, Inc. Bulacan,
2003
17. 7-7
EXAMPLE 1
The monthly incomes of recent MBA
graduates in a large corporation are
normally distributed with a mean of $2000
and a standard deviation of $200. What is
the Z value for an income of $2200? An
income of $1700?
For X=$2200, Z=(2200-2000)/200=1.
18. 7-8
EXAMPLE 1 continued
For X=$1700, Z =(1700-2000)/200= -1.5
A Z value of 1 indicates that the value of
$2200 is 1 standard deviation above the
mean of $2000, while a Z value of $1700 is
1.5 standard deviation below the mean of
$2000.
19. 7-9
Areas Under the Normal Curve
About 68 percent of the area under the
normal curve is within one standard
µ ± ofσ
deviation 1 the mean.
About 95 percent is within two standard
deviations of the mean. ± 2σ
µ
99.74 percent is within three standard
deviations of the mean.
µ ± 3σ
20. 7-10
Areas Under the Normal Curve
r a l i t r b u i o n : µ = 0 , σ2 = 1
0 . 4
Between:
1.68.26%
0 . 3 2.95.44%
3.99.74%
0 . 2
f ( x
0 . 1
. 0
µ µ + 2σ
- 5
µ − 2σ
µ + 3σ
x
µ − 3σ µ −1σ µ +1σ
Jonathan M. Pacurib
21. 7-11
The daily water usage per person in New
EXAMPLE 2
Providence, New Jersey is normally distributed
with a mean of 20 gallons and a standard
deviation of 5 gallons.
About 68% of the daily water usage per
person in New Providence lies between what
two values?
That is, about 68% of the daily water
usage will lie between 15 and 25 gallons.
µ ± 1σ = 20 ± 1(5).
22. 7-12
EXAMPLE 3 atthat a personuse less than
What is the probability
Providence selected random will
from New
20 gallons per day?
The associated Z value is Z=(20-20)/5=0. Thus,
P(X<20)=P(Z<0)=.5
What percent uses between 20 and 24 gallons?
The Z value associated with X=20 is Z=0 and with
X=24, Z=(24-20)/5=.8. Thus,
P(20<X<24)=P(0<Z<.8)=28.81%
23. 7-13 r a l i
EXAMPLE 3
t r b u i o n : µ = 0 ,
0 . 4
0 . 3
P(0<Z<.8)
=.2881
0 . 2
0<X<.8
f ( x
0 . 1
. 0
- 5
-4 -3 -2 -1 x
0 1 2 3 4
24. 7-14
EXAMPLE 3 continued
What percent of the population uses between
18 and 26 gallons?
The Z value associated with X=18 is Z=(18-
20)/5= -.4, and for X=26,
Z=(26-20)/5=1.2. Thus P(18<X<26)
=P(-.4<Z<1.2)=.1554+.3849=.5403
25. 7-15
Professor Mann has determined that the
EXAMPLE 4
final averages in his statistics course is
normally distributed with a mean of 72 and
a standard deviation of 5. He decides to
assign his grades for his current course such
that the top 15% of the students receive an
A. What is the lowest average a student
can receive to earn an A?
Let X be the lowest average. Find X such
that P(X >X)=.15. The corresponding Z
value is 1.04. Thus we have (X-72)/5=1.04,
or X=77.2
26. 7-16 r
EXAMPLE 4
a l i t r b u i o n : µ = 0 , σ2 = 1
0 . 4
0 . 3
Z=1.04
0 . 2
f ( x
0 . 1
. 0
15%
-
0 1 2 3 4
27. 7-17
The amount of tip the servers in an exclusive
EXAMPLE 5
restaurant receive per shift is normally
distributed with a mean of $80 and a standard
deviation of $10. Shelli feels she has provided
poor service if her total tip for the shift is less
than $65. What is the probability she has
provided poor service?
Let X be the amount of tip. The Z value
associated with X=65 is Z= (65-80)/10=
-1.5. Thus P(X<65)=P(Z<-1.5)=.5-.4332=.0668.
28. 7-18
The Normal Approximation to
Using the normal distribution (a continuous
the Binomial
distribution) as a substitute for a binomial
distribution (a discrete distribution) for large
values of n seems reasonable because as n
increases, a binomial distribution gets closer and
closer to a normal distribution.
The normal probability distribution is generally
deemed a good approximation to the binomial
probability distribution when n and n(1- ) are
both greater than 5.
π
π
29. 7-19
The Normal Approximation
continued
Recall for the binomial experiment:
There are only two mutually exclusive
outcomes (success or failure) on each
trial.
A binomial distribution results from
counting the number of successes.
Each trial is independent.
The probability is fixed from trial to trial,
and the number of trials n is also fixed.
30. Binomial Distribution for an n of 3
7-20
and 20,
where π =.50
n=3
n=20
0.4
0.3 0.2
0.15
P(x)
0.2
P(x) 0.1
0.1
0.05
0 0
0 1 2 3 2 4 6 8 10 12 14 16 18 20
number of occurences
number of occurences
31. 7-21
Continuity Correction Factor
Thevalue .5 subtracted or added, depending
on the problem, to a selected value when a
binomial probability distribution (a discrete
probability distribution) is being approximated
by a continuous probability distribution (the
normal distribution).
32. 7-22
EXAMPLE 6
A recent study by a marketing research
firm showed that 15% of American
households owned a video camera. A
sample of 200 homes is obtained.
Of the 200 homes sampled how many
would you expect to have video
cameras?
µ = nπ = (.15)(200) = 30
33. 7-23
What is the variance?
EXAMPLE 6
σ =nπ 1 − ) =( 30)(1− ) =25.5
2
( π .15
What is the standard deviation?
σ =
What is the.5 =5.0498 less than 40 homes
25 probability that
in the sample have video cameras? We need
P(X<40)=P(X<39). So, using the normal
approximation, P(X<39.5) P[Z
≈
(39.5-30)/5.0498] = P(Z 1.8812)
P(Z<1.88)=.5+.4699+.9699
≤ ≤ ≈
34. 7-24 µ σ2
EXAMPLE 6
r a l i t r b u i o n : = 0 , = 1
0 . 4
P(Z=1.88)
0 . 3
.5+.4699
=.9699
0 . 2
f ( x
0 . 1
Z=1.88
. 0
- 5
0 1 2 3 4
Jonathan M. Pacurib
