Transcript    

Converting to a Standard Normal Distribution
z
x
Think of me as the measure
of the distance from the
mean, measured in
standard deviations

Using Excel to Compute
Standard Normal Probabilities

Excel has two functions for computing probabilities
and z values for a standard normal distribution:
NORM S DIST is used to compute the cumulative
NORMSDIST
probability given a z value.
NORMSINV
NORM S INV is used to compute the z value
given a cumulative probability.
(The “S” in the function names reminds
us that they relate to the standard
normal probability distribution.)
Using Excel to Compute
Standard Normal Probabilities

Formula Worksheet
1
2
3
4
5
6
7
8
9
A
B
Probabilities: Standard Normal Distribution
P (z < 1.00)
P (0.00 < z < 1.00)
P (0.00 < z < 1.25)
P (-1.00 < z < 1.00)
P (z > 1.58)
P (z < -0.50)
=NORMSDIST(1)
=NORMSDIST(1)-NORMSDIST(0)
=NORMSDIST(1.25)-NORMSDIST(0)
=NORMSDIST(1)-NORMSDIST(-1)
=1-NORMSDIST(1.58)
=NORMSDIST(-0.5)
Using Excel to Compute
Standard Normal Probabilities

Value Worksheet
1
2
3
4
5
6
7
8
9
A
B
Probabilities: Standard Normal Distribution
P (z < 1.00)
P (0.00 < z < 1.00)
P (0.00 < z < 1.25)
P (-1.00 < z < 1.00)
P (z > 1.58)
P (z < -0.50)
0.8413
0.3413
0.3944
0.6827
0.0571
0.3085
Using Excel to Compute
Standard Normal Probabilities

Formula Worksheet
1
2
3
4
5
6
A
B
Finding z Values, Given Probabilities
z value with .10 in upper tail
z value with .025 in upper tail
z value with .025 in lower tail
=NORMSINV(0.9)
=NORMSINV(0.975)
=NORMSINV(0.025)
Using Excel to Compute
Standard Normal Probabilities

Value Worksheet
1
2
3
4
5
6
A
B
Finding z Values, Given Probabilities
z value with .10 in upper tail
z value with .025 in upper tail
z value with .025 in lower tail
1.28
1.96
-1.96
Example: Pep Zone
• Standard Normal Probability Distribution
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.
Pep
Zone
5w-20
Motor Oil
Example: Pep Zone
Pep
Zone
5w-20
Motor Oil

Standard Normal Probability Distribution
The store manager is concerned that sales are
being lost due to stockouts while waiting for an
order. It has been determined that demand during
replenishment lead time 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, P(x > 20).
Solving for Stockout Probability
Pep
Zone
5w-20
Motor Oil
Step 1: Convert x to the standard normal distribution
z
x

20  15

 .83
6
Thus 20 gallons sold during the
replenishment lead time would be
.83 standard deviations above the
average of 15.
Solving for Stockout Probability:
Step 2
Now we need to find the area under
the curve to the left of z = .83. This
will give us the probability that
x ≤ 20 gallons.
Pep
Zone
5w-20
Motor Oil
Example: Pep Zone
Pep
Zone
5w-20
Motor Oil
Cumulative Probability Table for
the Standard Normal Distribution

z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7
.7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
.
.
.
.
.
.
P(z <
.83)
.
.
.
.
.
Example: Pep Zone
Pep
Zone

Solving for the Stockout Probability
5w-20
Motor Oil
Step 3: Compute the area under the standard normal
curve to the right of z = .83.
P(z > .83) = 1 – P(z < .83)
= 1- .7967
= .2033
Probability
of a stockout
P(x > 20)
Example: Pep Zone
Pep
Zone

5w-20
Motor Oil
Solving for the Stockout Probability
Area = 1 - .7967
Area = .7967
= .2033
0
.83
z
Example: Pep Zone
Pep
Zone
5w-20
Motor Oil
If the manager of Pep Zone
wants the probability of a
stockout to be no more than
.05, what should the reorder
point be?
Example: Pep Zone
Pep
Zone
5w-20
Motor Oil

Solving for the Reorder Point
Area = .9500
Area = .0500
0
z.05
z
Example: Pep Zone
Pep
Zone

5w-20
Motor Oil
Solving for the Reorder Point
Step 1: Find the z-value that cuts off an area of .05
in the right tail of the standard normal
distribution.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732
.9750
.9756 .9761 .9767
We.9738
look.9744
up the
complement
.
.
.
.
. of the
. tail .area (1
. - .05. = .95).
.
Solving for the Reorder Point
Step 2: Convert z.05 to the corresponding value of x :
x    z.05  15  1.645(6)  24.87
Pep
Zone
5w-20
Motor Oil
Solving for the Reorder Point
Pep
Zone
5w-20
Motor Oil
So if we raising our reorder point
from 20 to 25 gallons, we reduce
the probability of a stockout from
about .20 to less than .05
Using Excel to Compute
Normal Probabilities

Excel has two functions for computing cumulative
probabilities and x values for any normal
distribution:
NORMDIST is used to compute the cumulative
probability given an x value.
NORMINV is used to compute the x value given
a cumulative probability.
Using Excel to Compute
Normal Probabilities
Formula Worksheet

1
2
3
4
5
6
7
8
Pep
Zone
5w-20
Motor Oil
A
B
Probabilities: Normal Distribution
P (x > 20) =1-NORMDIST(20,15,6,TRUE)
Finding x Values, Given Probabilities
x value with .05 in upper tail =NORMINV(0.95,15,6)
Using Excel to Compute
Normal Probabilities
5w-20
Motor Oil
Value Worksheet

1
2
3
4
5
6
7
8
Pep
Zone
A
B
Probabilities: Normal Distribution
P (x > 20) 0.2023
Finding x Values, Given Probabilities
x value with .05 in upper tail 24.87
Note: P(x > 20) = .2023 here using Excel, while our
previous manual approach using the z table yielded
.2033 due to our rounding of the z value.
Exercise 18, p. 261
The average time a subscriber reads the Wall Street Journal
is 49 minutes. Assume the standard deviation is 16
minutes and that reading times are normally distributed.
a) What is the probability a subscriber will spend at least one
hour reading the Journal?
b) What is the probability a reader will spend no more than 30
minutes reading the Journal?
c) For the 10 percent who spend the most time reading the
Journal, how much time do they spend?
Exercise 18, p. 261
a) Convert x to the standard normal distribution:
x
60  49
 .6875

16
Thus one who read 560 minutes would be .69 from
the mean. Now find P(z ≤ .6875). P(z ≤ .69)= .7549.
Thus P(x > 60 minutes) = 1 - .7549 = .2541.
z

b) Convert x to the standard normal distribution
z
30  49
 1.19
16
P(x ≤ 30 minutes)
P( z  1.19)  1  P( z  1.19)
 1  .8830  .117
Red-shaded
area is equal to
blue shaded
area
Thus:
P(x < 30 minutes) = .117
-1.19
0
1.19
z
Exercise 18, p. 261
(c)
x    z.10 
49  1.285(16)  70 minutes or more