Transcript QBM117 Business Statistics

QBM117
Business Statistics
Probability and Probability Distributions
The Normal Distribution continued
1
Objectives
•
To learn how to use the Z tables in reverse to find
the value of Z corresponding to a known probability.
•
To learn how to use the Z tables in reverse to find
the value of X corresponding to a known probability.
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Finding Values That Correspond to
Known Probabilities
• Start by looking on the inside of the table for the
known probability.
• Then move to the outside of the table to determine
the associated z value.
• Then back transform to obtain the associated x value.
X    Z
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Example 1
An area of 0.4370 lies under the standard normal
curve between the mean and a given positive z
score. What is the value of that z score?
We want to find z* such that
P(0  Z  z* )  0.4370
0.4370
0
z*
Z
The z value corresponding to the area of 0.4370 is
z*  1.53
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Example 2
An area of 0.25 lies under the standard normal curve
between the mean and a given positive z score. What
is the value of that z score?
We want to find z* such that
P(0  Z  z* )  0.25
0.25
0 z*
Z
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To find z* we search the table for the probability 0.25.
We don’t find this probability but we find two that are
close: 0.2486 and 0.2517.
0.25 is closer to 0.2486 than it is to 0.2517.
And so we look up the z value associated with
0.2486, which is z*  0.67
Example 3
An area of 0.05 lies under the standard normal curve
above a given positive z score. What is the value of
that z score?
We want to find z* such that
P( Z  z* )  0.05
0.05
0
z*
Z
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Hence we want to find z* such that
P(0  Z  z* )  0.5  0.05
 0.45
0.45
0.05
0
z*
Z
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To find z* we search the table for the probability 0.45.
We don’t find this probability but we find two that are
close: 0.4495 and 0.4505
0.45 is exactly half way between 0.4495 and 0.4505.
And so we look up the z value associated with both
probabilities and average them.
The Z values associated with these probabilities are
1.64 and 1.65.
The average of these values is 1.645, hence
z*  1.645
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Exercise 1
Find the value z* for which
a.
P( Z  z* )  0.15
b.
P( Z  z* )  0.80
c.
P( z*  Z  z* )  0.90
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Example 3
Scores of an aptitude test given by a training
department of a large company are normally
distributed with a mean of 75 points and a standard
deviation of 5 points.The company has decided that
people who score in the bottom 10% of the test
scores will not receive any additional job training. If
there are to be layoffs, these people will be among
the first to be cut. What cut-off score on the test
should the company use?
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Let X = the score on the aptitude test
2
X ~ N (75,5 )
We want to find the value of X that has 10% below it.
0.10
x*
75
X
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First we find the value of Z that has 10% below it and
then we back transform to find the corresponding
value of X.
We want to find z* such that P( Z  z* )  0.10
0.40
0.10
z*
Z
0
Using the Z tables we find
z*  1.28
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We now need to back transform to find x*
x*    z*
 75  (1.28)  5
 68.6
Therefore the company should set the cut-off at 68.6
points.
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Example 3 continued
The company is planning to give extra training to
employees who score in the top 2% of those taking
the test. The company would like to identify the score
to use as the cut-off point.
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Recall that X = the score on the aptitude test
2
X ~ N (75,5 )
We want to find the value of X that has 2% above it.
0.02
75
x*
X
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First we need to find the value of Z that has 2%
above it.
0.48
0.02
0
From the Z tables we find
Hence
z*
z*  2.05
Z
x*  75  2.05  5
 85.25
Therefore the cut-off should be 82.25 points.
Exercise 2
The marks for a first year statistics exam are
normally distributed with a mean of 72 and a
standard deviation of 14.
Suppose the lecturer wants to assign High
Distinctions to the top 15% and fail the bottom 20%.
a. What is the cut-off score for a HD?
b. What is the pass mark?
Exercise 3
The number of pages printed before replacing the
cartridge in a laser printer is normally distributed with
a mean of 11500 pages and a standard deviation of
800 pages. The manufacturer wants to provide
guidelines to potential customers advising them the
minimum number of pages they can expect from
each cartridge. How many pages should it advertise if
the company wants to be correct 99% of the time?
Exercise 4
The Rural Bank is reviewing its service charges and
interest-paying policies on cheque accounts. The
bank has found that the average daily balance on
personal cheque accounts is normally distributed
with a mean of $550.00 and a standard deviation of
$150.00.
a. What percentage of personal cheque account
customers carry average daily balances below
$200?
b. The bank is considering paying interest to customers
carrying average daily balances in excess of a
certain amount. If the bank does not want to pay
interest to more than 8% of its customers, what is
the minimum average daily balance it should be
willing to pay interest on?
Exercises
• 5.33
• 5.79
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