Chapter 6 Part 2

1
Chapter 6 Part 2

• Continuous Probability Distributions (Continued)

2
Normal Distribution

• Since there is a common thread between all normal
  distributions (with respect to z scores), we can
  use a single source to determine area under the
  normal curves.
• The Standard Normal Distribution is used to
  accomplish this task.

3
Standard Normal Distribution

• If x is normally distributed with mean ? and
  standard deviation ?, then
• x is the value of the corresponding
• z score.
• is normally distributed with mean 0 and standard
  deviation 1, and is called the standard normal
  distribution.
• Clearly, the z score of the mean ? is 0, since
  plugging ? in the above z score formula for x
  yields a z value of zero.
• We can think of z as a measure of the number of
  standard deviations x is from ?.

4
The Standard Normal Table

• The Standard Normal Table is in the front of your
  book. This table gives the
• area under a normal curve to the left of any z
  score value that you look
• up.
• If you want to know the area under the normal
  curve that is to the left of 1.57 , go to the row
  for 1.0 and the column for .07 and the row and
  column will meet at the corresponding area. On
  the following slide we see that the probability
  that a random variable takes on a value less than
  1.57 is 0.9418.

P(zlt 1.57)
z
1.57
5
Standard Normal Table
P(Zlt1.57)
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The Standard Normal Table

• If you want to know the area under the normal
  curve to the left of 1.90, go to the row for 1.9
  and the column for 0.00 and the row and column
  will meet at the corresponding area. From the
  table we see that the probability that a random
  variable takes on a value less than 1.90 is
  0.9713.

z
1.90
7
Standard Normal Table
P(Zlt1.90)
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Areas under Normal Curves and Negative z Scores

• Since the Normal Curve is symmetric about the
  mean, the area under the curve between the mean
  and z is the same as the area under the curve
  between the mean and z.
• Thus area under the normal curve between z0 (the
  mean) and a
• z score of -0.50 or P(-0.50 ? z ? 0) P(0 ? z ?
  0.5) 0.1915.
• Also P(0 ? z ? 0.95) P(-0.95 ? z ? 0) 0.3289.

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Practice Problems

• Determine P(z lt -1)
• Determine P(z lt 1)
• Determine P(z ? -1)
• Determine P(z ? 1)
• Determine P(0 ? z ? 2)
• Determine P(-1.5 ? z ? 0)
• a. Determine P(-1 ? z ? 1)
• b. Determine P(-2 ? z ? 2)
• c. Determine P(-3 ? z ? 3)
• d. Determine P(-2.53 ? z ? 2.53)

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Practice Problem 1
Calculating P(z lt -1)
Steps 1-3
z
Step 4
From chart, P(zlt-1)0.1587
z
11
Practice Problem 2
Calculating P(z lt 1)
Steps 1-3
z
Step 4
From chart, P(zlt1)0.8413
z
z
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Practice Problem 3
Calculating P(z ? -1)
Step 4
a.) From chart, P(zlt-1)0.1587
Steps 1-3
z
z

• b.) Since area under curve is 1
• P(zgt-1) 1-P(zlt-1)
• .8413

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Practice Problem 4
Calculating P(z ? 1)
Step 4
Steps 1-3
a.) From chart, P(zlt1) 0.8413
z
z

• b.) Since area under curve is 1
• P(zgt1) 1-P(zlt1)
• 0.1587

Note that P(zgt1) equals P(zlt-1)
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Practice Problem 5
Calculating P(0ltz lt 2)
Step 4
Steps 1-3
a.) From chart, P(zlt2) 0.9772
z
z

• b.) Since area under curve is 1 and half the area
  under the curve is 0.5
• P(0ltz lt 2) 0.9772-0.5
• 0.4773

FINISH THE OTHER PRACTICE PROBLEMS
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Steps for Determining Normal Probabilities

• Formulate your problem in terms of x.
• Restate your problem in terms of the
  corresponding z values.
• Draw you Normal curve and shade in the region
  under the curve in the interval of interest.
• Use the standard normal tables to find the
  indicated area under the normal curve.
• NOTE The area under the normal curve calculated
  in step 4 equals
• P(altxltb)

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Example 6.3

• Pep Zone sells auto parts and supplies including
  a
• popular multi-grade motor oil. When the stock of
  this
• oil drops to 20 gallons, a replenishment order is
  placed.
• The store manager is concerned that sales are
  being
• lost due to stockouts while waiting for an order.
  It has
• been determined that demand is normally
• distributed with a mean of 15 gallons and a
  standard
• deviation of 6 gallons.
• The manager would like to know the probability
  of a
• stockout- defined as P(x gt 20).

Anderson, Sweeney, and Williams
17

• Following the steps for determining normal
  probabilities
• We are trying to find P(x gt 20).
• Restate the problem in terms of z.
• The z score for the value 20 is
• z
• Thus P(xgt20) can be restated as P(zgt.83)

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3. Draw the curve and shade the appropriate
region.
x
15
20
z
.83
0
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4. Determine the shaded area.
The Standard Normal table shows an area of 0.7967
for the region less than z 0.83 (See next
slide). Since the area under the entire curve
equals 1.0,
P(xgt20)1.0-0.7967.2033
0.7967
.2033
x
15
20
z
.83
0
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Standard Normal Probability Distribution
Cumulative Probability Table for the Standard
Normal Distribution
P(z lt .83)
Anderson, Sweeney, and Williams
P(z lt .83)0.7967
0.7967
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Example 6.4

• The grades on a statistics midterm exam were
  normally distributed with a
• mean of 72 and a standard deviation of 8.
• What is the proportion of students received a B
  grade.
• b. What is the probability that a randomly
  selected student received between a 65 and 85?
• c. What is the proportion of students that failed
  the exam?

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Example 6.4a

• Following the steps for determining normal
  probabilities
• We are trying to find P(80 lt x lt 90).
• This can be restated as P(1lt z lt 2.25)

Note Recall that for a continuous random
variable the probability that a random variable
equals a specific value equals 0. Thus inserting
? symbol above would not change the answer to
the above stated problem.
The z score for 90 is
Since the z score for 80 is

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3. Draw the curve and shade the appropriate
region.
x
90
72
80
z
2.25
1
0
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4. Determine the shaded area

• The Standard Normal table shows an area of .8413
  for the region less than z 1.00.
• It also shows an area of .9878 for the region
  between z 2.25.

P(80gtxgt90)0.9878-0.8413 0.1465
.8413
Thus 14.65 of the students received a B
.9878
x
90
72
80
z
1
0
2.25
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Example 6.4b
Find P(65ltxlt85). This can be restated as P(-.88 lt
z lt 1.63). The Standard Normal table shows an
area of 0.1894 for the region less than z -0.88
and an area of .9484 for the region less than z
1.63.
P(65ltxlt85)0.9484-0.1894 .759
.9484
0.1894
72
85
65
x
-.88
0
1.63
z
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Example 6.4c
Find P(x lt 60). This can be restated as P(z lt
-1.5). The Standard Normal table shows an area
of 0.0668 for the region less than z -1.5.
P(xlt60) .0668
.3106
So Approximately 6.68 of the students will fail
0.668
72
60
x
-1.5
0
z
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Example 6.3 Revisited (Inverse Normal Problem)

• Refer to slide 14. If the manager of Pep Zone
  wants the probability of a
• stockout to be no more than .05, what should the
  reorder point be?
• Lets call the reorder point x?. From above we
  know that P(x gt x?) 0.05.

We drew the graph below because we knew that a
stockout means that the demand exceeds the
reorder point. Thus the interval of interest is
all values that exceed the reorder point. We
shaded the area in this interval.
15
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Standard Normal Probability Distribution
Solving for the Reorder Point
Find the z-value that cuts off an area of .05 in
the right tail of the standard normal
distribution.
We look up the complement of the tail area (1 -
.05 .95)
Anderson, Sweeney, and Williams
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Example 6.3 Revisited (Inverse Normal Problem)

• If z1.645, then
  .
• Thus, x 24.87.

.05
Since the prob of falling in interval is .05
.9500
x?
x
15
z
0
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• The End