Download Normal Distribution

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Chapter
The Normal Probability
Distribution
THIS CHAPTER’S GOALS
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TO LIST THE CHARACTERISTICS OF THE
NORMAL DISTRIBUTION.
TO DEFINE AND CALCULATE Z VALUES.
TO DETERMINE PROBABILITIES
ASSOCIATED WITH THE STANDARD
NORMAL DISTRIBUTION.
TO USE THE NORMAL DISTRIBUTION TO
APPROXIMATE THE BINOMIAL
DISTRIBUTION.
CHARACTERISTICS OF A NORMAL
PROBABILITY DISTRIBUTION
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 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 this
center point, and the other half is below it.
 The normal probability distribution is
symmetrical about its mean.
 It is asymptotic - the curve gets closer and closer
to the x-axis but never actually touches it.
CHARACTERISTICS OF A NORMAL DISTRIBUTION
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Normal curve is symmetrical
two halves identical
-
Tail
Tail
0.5
Theoretically, curve
extends to - infinity
0.5
Mean, median, and
mode are equal
Theoretically, curve
extends to + infinity
Normal Distributions with Equal Means
but Different Standard Deviations.
s = 3.1
s = 3.9
s = 5.0
m = 20
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Normal Probability Distributions with Different
Means and Standard Deviations.
m = 5, s = 3
m = 9, s = 6
m = 14, s = 10
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THE STANDARD NORMAL
PROBABILITY DISTRIBUTION
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 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 m,
divided by the population standard deviation, s.
X

m
Z= s
 The z-value is the number of standard deviations
X is from the mean.
EXAMPLE
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 The monthly incomes of recent MBA graduates
in a large corporation are normally distributed
with a mean of $2,000 and a standard deviation
of $200. What is the z value for an income X of
$2,200? $1,700?
 For X = $2,200 and since z = (X - m)/s, then
z = (2,200 - 2,000)/200 = 1.
 A z value of 1 indicates that the value of $2,200 is
1 standard deviation above the mean of $2,000.
EXAMPLE (continued)
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 For X = $1,700 and since z = (X - m)/s, then
z = (1,700 - 2,000)/200 = -1.5.
 A z value of -1.5 indicates that the value of
$2,200 is 1.5 standard deviation below the mean
of $2,000.
 How might a corporation use this type of
information?
AREAS UNDER THE NORMAL CURVE
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 About 68 percent of the area under the normal
curve is within plus one and minus one standard
deviation of the mean. This can be written as m
± 1s.
 About 95 percent of the area under the normal
curve is within plus and minus two standard
deviations of the mean, written m ± 2s.
 Practically all (99.74 percent) of the area under
the normal curve is within three standard
deviations of the mean, written m ± 3s.
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Between:
1. 68.26%
2. 95.44%
3. 99.97%
m+3s
m2s m1s
m
m+1s m+2s
m+3s
P(z)=?
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 A typical need is to determine the probability of
a z-value being greater than or less than some
value.
 Tabular Lookup (Appendix D, page 474)
 EXCEL Function
=NORMSDIST(z)
EXAMPLE
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 The daily water usage per person in Toledo,
Ohio 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 Toledo lies between what two values?
EXAMPLE
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 The daily water usage per person in Toledo (X),
Ohio is normally distributed with a mean of 20
gallons and a standard deviation of 5 gallons.
 What is the probability that a person selected at
random will use less than 20 gallons per day?
 What is the probability that a person selected at
random will use more than 20 gallons per day?
EXAMPLE (continued)
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 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 = 0.8
P(20
< X < 24) = P(0 < z < 0.8) = 0.2881=28.81%
 What percent uses between 16 and 20 gallons?
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P(0 < z < 0.8)
= 0.2881
0.8
EXAMPLE (continued)
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 What is the probability that a person selected at
random uses more than 28 gallons?
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P(z > 1.6) =
0.5 - 0.4452 =
0.0048
Area =
0.4452
1.6
z
EXAMPLE (continued)
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 What percent uses between 18 and 26 gallons?
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Total area =
0.1554 + 0.3849 =
0.5403
Area =
0.1554
Area =
0.3849
 .4
1.2
z
EXAMPLE (continued)
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 How many gallons or more do the top 10% of
the population use?
 Let X be the least amount. Then we need to find
Y such that P(X  Y) = 0.1 To find the
corresponding z value look in the body of the
table for (0.5 - 0.1) = 0.4. The corresponding z
value is 1.28 Thus we have (Y - 20)/5 = 1.28,
from which Y= 26.4. That is, 10% of the
population will be using at least 26.4 gallons
daily.
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(Y - 20)/5 = 1.28
Y= 26.4
0.4
0.1
1.28
z
EXAMPLE
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 A professor has determined that the 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 must receive to earn an A?
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(Y - 72)/5 = 1.04
Y = 77.2
0.35
0.15
1.04
z
EXAMPLE
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 The amount of tip the waiters in an exclusive
restaurant receive per shift is normally
distributed with a mean of $80 and a standard
deviation of $10. A waiter feels he has provided
poor service if his total tip for the shift is less
than $65. Based on his theory, what is the
probability that he has provided poor service?
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Area =
0.5 - 0.4332 =
0.0668
Area =
0.4332
- 1.5
z
THE NORMAL APPROXIMATION TO
THE BINOMIAL
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 Using the normal distribution (a continuous
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.
 When to use the normal approximation?
 The normal probability distribution is generally
deemed a good approximation to the binomial
probability distribution when np and n(1 - p) are
both greater than 5.
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Binomial Distribution with n = 3 and p = 0.5.
P(r)
0.5
0.4
0.3
0.25
r
0
1
2
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Binomial Distribution with n = 5 and p = 0.5.
P(r)
r
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Binomial Distribution with n = 20 and p = 0.5.
P(r)
Observe
the
Normal
shape.
r
THE NORMAL APPROXIMATION
(continued)
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 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 p is fixed from trial to trial, and
the number of trials n is also fixed.
CONTINUITY CORRECTION
FACTOR
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 The value 0.5 subtracted or added, depending on
the problem, to a selected value when a binomial
probability distribution, which is a discrete
probability distribution, is being approximated
by a continuous probability distribution--the
normal distribution.
 The basic concept is that a slice of the normal
curve from x-0.5 to x+0.5 is approximately equal
to P(x).
EXAMPLE
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 A recent study by a marketing research firm
showed that 15% of the homes had a video
recorder for recording TV programs. A sample
of 200 homes is obtained. (Let X be the number
of homes).
 Of the 200 homes sampled how many would you
expect to have video recorders?
 m = np (0.15)(200) = 30 & n(1 - p) = 170
 What is the variance?
 s2 = np(1 - p) = (30)(1- 0.15) = 25.5
EXAMPLE (continued)
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 What is the standard deviation?
 s = (25.5) = 5.0498.
 What is the probability that less than 40 homes
in the sample have video recorders?
 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) = 0.5 + 0.4699 = 0.9699
Why did I use 39.5 ? ...
How would you calculate P(X=39) ?
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P(z  1.88) =
0.5 + 0.4699 =
0.9699
0.5
0.4699
1.88
z
EXAMPLE (continued)
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 What is the probability that more than 24 homes
in the sample have video recorders?
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P(z  -1.09) =
0.5 + 0.3621 =
0.8621.
0.3621
-1.09
0.5
z
EXAMPLE (continued)
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 What is the probability that exactly 40 homes in
the sample have video recorders?
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P(1.88  z  2.08)
= 0.4812 - 0.4699
= 0.0113
1.88
2.08
0.4699
0.4812
z