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Week 4
Dr. Jenne Meyer
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In order of last name
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Miracle
Brandi
Sandy
LeAndrew
Bren
Christine
Brandon
Maria
Rose
Monique
Michelle
Jodi
Discrete Variable – each
value of X has its own
probability P(X).
 Continuous Variable –
events are intervals and
probabilities are areas
underneath smooth curves.
A single point has no
probability.
 Total area under curve = 1
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Defined by two parameters, m and s
Almost all area under the normal curve is included
in the range m – 3s < X < m + 3s
All normal distributions have the same
shape but differ in the axis scales.
m = 42.70mm
s = 0.01mm
Diameters of golf balls
McGraw-Hill/Irwin
m = 70
s = 10
CPA Exam Scores
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
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Since for every value of m and s, there is
a different normal distribution, we
transform a normal random variable to a
standard normal distribution with m = 0
and s = 1 using the formula:
x–m
z=
s
Appendix allows you to find the
area under the curve from 0
to z.
Now find P(Z < 1.96):
.5000
.5000 - .4750 = .0250
Now find P(-1.96 < Z < 1.96).
Due to symmetry, P(-1.96 < Z) is the same as P(Z
< 1.96).
.9500
So, P(-1.96 < Z < 1.96) = .4750 + .4750 = .9500
or 95% of the area under the curve.
Some important Normal areas:
Suppose John took an economics exam and scored 86
points. The class mean was 75 with a standard deviation
of 7. What percentile is John in (i.e., find P(X < 86)?
86 – 75 = 11/7 = 1.57
x
–
m
zJohn =
=
7
s
So John’s score is 1.57 standard deviations about the mean.
Suppose John took an economics exam and scored 86
points. The class mean was 75 with a standard deviation
of 7. What percentile is John in (i.e., find P(X < 86)?
Suppose John took an economics exam and scored 86
points. The class mean was 75 with a standard deviation
of 7. What percentile is John in (i.e., find P(X < 86)?
normal distribution
p(lower) p(upper)
z x mean std.dev
.9420
.0580 1.57 86
75
7
Suppose John took an economics exam and scored 86
points. The class mean was 75 with a standard deviation
of 7. What percentile is John in (i.e., find P(X < 86)?
John is approximately in the 94th percentile
For example, let m = 2.040 cm and s = .001 cm,
what is the probability that a given steel bearing
will have a diameter between 2.039 and
2.042cm?
 In other words, P(2.039 < X < 2.042)
 Excel only gives left tail areas, so break the
formula into two, find P(X < 2.039) and
P(X < 2.042), then subtract them to find the
desired probability:
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P(X < 2.042) = .9773
P(X < 2.039) = .1587
P(2.039 < X < 2.042) = .9773 - .1587 = .8186 or 81.9%
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suppose we wanted the probability of
selecting a foreman who earned less than
$1,100. In probability notation we write this
statement as P(weekly income < $1,100).

suppose we wanted the probability of
selecting a foreman who earned less than
$1,100. In probability notation we write this
statement as P(weekly income < $1,100).
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=.8413
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suppose we wanted the probability of
selecting a foreman who earned less than
$1,100. In probability notation we write this
statement as P(weekly income < $1,100).
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=.8413
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The mean of a normal probability distribution is 500; the
standard deviation is 10.
a. About 68 percent of the observations lie between what
two values?
b. About 95 percent of the observations lie between what
two values?
c. Practically all of the observations lie between what two
values?
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The mean of a normal probability distribution is 500; the
standard deviation is 10.
a. About 68 percent of the observations lie between what
two values?
b. About 95 percent of the observations lie between what
two values?
c. Practically all of the observations lie between what two
values?
a. 490 and 510, found by 500 +/- 1(10).
b. 480 and 520, found by 500 +/- 2(10).
c. 470 and 530, found by 500 +/- 3(10).
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A normal distribution has a mean of 50 and a standard
deviation of 4.
a. Compute the probability of a value between 44.0 and 55.0.
b. Compute the probability of a value greater than 55.0.
c. Compute the probability of a value between 52.0 and 55.0.
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a. 0.8276: First find z -1.5, found by (44 - 50)/4 and
z = 1.25 = (55 - 50)/4. The area between -1.5 and 0 is
0.4332 and the area between 0 and 1.25 is 0.3944, both
from Appendix D. Then adding the two areas we find
that 0.4332 + 0.3944 = 0.8276.
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b. 0.1056, found by 0.5000 - 0.3994, where z = 1.25.
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c. 0.2029: Recall that the area for z = 1.25 is 0.3944, and
the area for z = 0.5, found by (52 - 50)/4, is 0.1915.
Then subtract 0.3944 - 0.1915 and find 0.2029.
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Next weeks assignments.
 Textbook Assignment: Complete Chapter 6 problems 17,
18, 19 (a, b, c), 20
 Textbook Assignment: Complete Chapter 7 problems 11, 13
 Moved Final Checkpoint to week 6
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