8- 1 Chapter Six The Normal Probability Distribution GOALS : When you have completed this chapter, you will be able to: •List the characteristics of the normal probability distribution. •Define and calculate z values. •Determine the probability an observation will lie between two points using the standard normal distribution. •Determine the probability an observation will be above or below a given value using the standard normal distribution. •Compare two or more observations that are on different probability distributions.
Transcript
8- 1
Chapter Six
The Normal Probability Distribution
GOALS : When you have completed this chapter, you will be able to:
•List the characteristics of the normal probability distribution.
•Define and calculate z values.
•Determine the probability an observation will lie between two points
using the standard normal distribution.
•Determine the probability an observation will be above or below a
given value using the standard normal distribution.
•Compare two or more observations that are on different probability
distributions.
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Characteristics of a Normal Probability Distribution
The normal curve is ___________ and has a single peak at the
exact center of the distribution.
The arithmetic of the distribution are
equal and located at the peak. Thus half the area under the curve
is above the mean and half is below it.
The normal probability distribution is symmetrical about its mean.
The normal probability distribution is asymptotic.That is the curve
gets closer and closer to the X-axis but never actually touches it.
bell-shaped
mean, and mode median,
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- 5
0 . 4
0 . 3
0 . 2
0 . 1
. 0
x
f(
x
l i : ,
Characteristics of a Normal Distribution
Mean, median & mode are equal
Normal curve is symmetrical
Theoretically, Curve
extends to infinity
a
Left = Right
Skip 2 slides
8- 4
The Standard Normal Probability Distribution
The standard normal distribution is a normal distribution
with a mean of and a standard deviation of
It is also called the z distribution.
σ
μx z
A z-value is the distance between a selected value,
designated X, and the population mean , divided by the
population standard deviation, σ . The formula is:
0 1.
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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:
XZ
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.
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EXAMPLE 1
The bi-monthly starting salaries of recent MBA graduates
follows the normal distribution with a mean of $2,000 and a
standard deviation of $200. What is the z-value for a salary of
$2,200? What is the z-value for a salary of $1,700?
00.1200$
000,2$200,2$
σ
μXz
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Comparing ___ and ___ units
Z
2000
1.00
2200 X ( = 2000, = 200)
( = 0, = 1)
ZX
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EXAMPLE 1 continued
A z-value of 1 indicates that the value of $2,200 is one standard
deviation above the mean of $2,000.
A z-value of –1.50 indicates that $1,700 is 1.5 standard deviation
below the mean of $2000.
Xz
What is the z-value of $1,700.
50.1200$
000,2$700,1$
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Areas Under the Normal Curve
About 68 percent of the area under the normal curve is within one standard deviation of the mean.
1 σ (0.3413x2=0.6826)
About 95 percent is within two standard deviations of the mean.
2 σ (0.4772x2=0.9544)
Practically all is within three standard deviations of the mean.
3 σ (0.4987x2=0.9974)
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EXAMPLE 2
The daily water usage per person in New Providence,
New Jersey is normally distributed with a mean of 20
gallons and a standard deviation of 5 gallons. About
68 percent of those living in New Providence will
use how many gallons of water?
About ____ of the daily water usage will lie between
15 and 25 gallons.
68%
Z
20
1.00
25 X15
-1.0
68% of the data
8- 11EXAMPLE 3
What is the probability that a person from New Providence selected
at random will use between 20 and 24 gallons per day?
Xz
80.05
2024
Xz
00.05
2020
The area under a normal curve between a z-value of 0 and a z-value of 0.80 is
0.2881.
We conclude that 28.81 percent of the residents use between 20 and 24 gallons of
water per day.
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P(20<X<24)
Z
20
0.80
24 X
Area=0.2881
=P(0<Z<0.8)=0.2881
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EXAMPLE 3 continued
What percent of the population use between 18 and 26 gallons per day?
40.05
2018
Xz
20.15
2026
Xz
The area associated with a z-value of –0.40 is 0.1554.
The area associated with a z-value of 1.20 is 0.3849.
Adding these areas, the result is 0.5403.
We conclude that 54.03 percent of the residents use between
18 and 26 gallons of water per day.
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Pg 239 Q18b(Textbook-Statistical Techniques in Business and Economics)
A normal population has a mean of 80 and a Std deviation of 14.
b) Compute the probability a value of 75 or less?
80 X75
μ=80 σ=14
Less : LHS
36.014
8075Xz
P(X ≤75)
= P(Z ≤ -0.36)
= 0.5-0.1406
= 0.3594
0.1406LHS of 75
-0.36 0 Z
8- 15Pg 185 Q17b(Textbook-Statistical Techniques in Business and Economics)
A normal population has a mean of 50 and a Std deviation of 4.
b) Compute the probability a value greater than 55?
50 55 X
µ=50 σ=4
Greater : RHS of 55
25.14
5055Xz
P(X >55)
= P(Z >1.25)
= 0.5-0.3944 (half of curve is 0.5)
= 0.1056
0.3944
0 1.25 z
+ 3 σ
8- 16Pg 185 Q17a(Textbook-Statistical Techniques in Business and Economics)
A normal population has a mean of 50 and a std deviation of 4.
a) Compute the probability a value between 44 and 55?
50x
55
µ =50 σ=4
25.14
5055Xz
P(44<X <55)
= P(-1.5<Z<1.25)
= 0.4332+0.3944
=0.8276
5.14
5044Xz
44
8- 17Pg 185 Q17c(Textbook-Statistical Techniques in Business and Economics)
A normal population has a mean of 50 and a std deviation of 4.
c) Compute the probability a value between 52 and 55?
50 x
μ=50 σ=4
25.14
5055Xz
P(52<X <55)
= P(0.5<Z<1.25)
=0.3944- 0.1915
=0.2029
5.04
5052Xz
52 55
0 0.5 1.25 z
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Pg 201 Q54
Fast Service Truck Lines uses the Ford Super Duty
F750. Management made a study of the maintenance
costs and determine the number of miles travelled
during the year .The mean of the distribution was
60,000 and the Std deviation 2,000 miles.
a) What percent of the Ford Super Duty F750 logged 65,200
miles or more?
b) What percent of the truck logged more than 57,060 but
less than 58,280 miles?
c) What percent of the Ford Super Duty F750 travelled
62,000 miles or less during the year?
d)Is it reasonable to conclude that the trucks were driven
more than 70,000 miles?
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Pg 201 Q54
a) What percent of the Ford Super Duty F750. logged 65,200
miles or more?
μ=60000 σ=2000
60000X
65,200
More : RHS
6.22000
6000065200Xz
P(X ≥65200)
= P(Z ≥ 2.6)
= 0.5-0.4953
=0.0047
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Pg 201 Q54
b) What percent of the truck logged more than 57,060 but less
than 58,280 miles?
μ=60000 σ=2000
60000X
86.02000
6000058280z
47.12000
6000057060z
2
1
P(57,060 <X < 58,280)
= P(-1.47<z <-0.86)
= 0.4292-0.3051
=0.1241
57,060 58,280
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Pg 201 Q54
c) What percent of the Ford Super Duty F750 logged 62,000
miles or less during the year?
μ=60000 σ=2000
60000
Z
62,000
Less : LHS
12000
6000062000z
P(X ≤62000)
= P(Z ≤1)
= 0.5+0.3413
=0.8413
0.34130.5
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Pg 2501Q54
d)Is it reasonable to conclude that the trucks were driven
more than 70,000 miles?
60000
0
X
Z
70,000
5
No, it is not reasonable to
conclude that the trucks were
driven more than 70,000 miles?
5000,2
000,60000,70z
P(x>70000)
= P(Z>5)
= 0.5-0.5
= 0
Z >3.9, Area=0.5
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Pg 201 Q54
a) 0.0047
b) 0.1241
c) 0.8413
d) No, because 5000,2
000,60000,70z
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EXAMPLE 4
Pg 202 Q 66 (Textbook-Statistical Techniques in Business
and Economics)
The price of shares of Bank of Florida at the end of trading
each day for the last year followed the normal distribution.
Assume there were 240 trading days in the year. The mean
price was $42 per share and the standard deviation was
$2.25 per share.
a) What percent of the days was the price over $45.00? How
many days would you estimate?
b) What percent of the days was the price between $38.00 and
$40.00?
c) What was the stock’s price on the highest 15% of days?
