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: is used to compute the cumulative probability given a z value.
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
A B
1
2
Probabilities: Standard Normal Distribution
3 P ( z < 1.00) =NORMSDIST(1)
4 P (0.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(0)
5 P (0.00 < z < 1.25) =NORMSDIST(1.25)-NORMSDIST(0)
6 P (-1.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(-1)
7 P ( z > 1.58)
8 P ( z < -0.50)
9
=1-NORMSDIST(1.58)
=NORMSDIST(-0.5)


Using Excel to Compute
Standard Normal Probabilities
Value Worksheet
A B
1
2
Probabilities: Standard Normal Distribution
3 P ( z < 1.00) 0.8413
4 P (0.00 < z < 1.00) 0.3413
5 P (0.00 < z < 1.25) 0.3944
6 P (-1.00 < z < 1.00) 0.6827
7 P ( z > 1.58)
8 P ( z < -0.50)
9
0.0571
0.3085


Using Excel to Compute
Standard Normal Probabilities
Formula Worksheet
A B
1
2
Finding z Values, Given Probabilities
3 z value with .10 in upper tail =NORMSINV(0.9)
4 z value with .025 in upper tail =NORMSINV(0.975)
5 z value with .025 in lower tail =NORMSINV(0.025)
6

Using Excel to Compute
Standard Normal Probabilities
 Value Worksheet
A B
1
2
Finding z Values, Given Probabilities
3 z value with .10 in upper tail 1.28
4 z value with .025 in upper tail 1.96
5 z value with .025 in lower tail -1.96
6

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

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).

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.

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

Example: 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
5w-20
Motor Oil

Solving for the Stockout Probability
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

Solving for the Stockout Probability
Area = .7967
Area = 1 - .7967
= .2033
5w-20
Motor Oil z
0 .83

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

Solving for the Reorder Point
Area = .9500
Area = .0500
5w-20
Motor Oil
0 z
.05
z

Example: Pep Zone

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.
5w-20
Motor Oil 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
.
.
.
.
.
.
.
.
.
.

Step 2 : Convert z
.05
to the corresponding value of x : x
   z
.
05
 
15

1 .
645 ( 6 )

24 .
87
5w-20
Motor Oil

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

 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
5w-20
Motor Oil
A B
1
2
3
4
Probabilities: Normal Distribution
P ( x > 20) =1-NORMDIST(20,15,6,TRUE)
5
6
Finding x Values, Given Probabilities
7 x value with .05 in upper tail =NORMINV(0.95,15,6)
8


Using Excel to Compute
Normal Probabilities
Value Worksheet
5w-20
Motor Oil
A B
1
2
3
4
Probabilities: Normal Distribution
P ( x > 20) 0.2023
5
6
Finding x Values, Given Probabilities
7 x value with .05 in upper tail 24.87
8
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.

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?

a) Convert x to the standard normal distribution: z
 x




60

49
16

.
6875
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 . b) Convert x to the standard normal distribution z

30

49
16
 
1 .
19

Red-shaded area is equal to blue shaded area
P(x ≤ 30 minutes)
P ( z

1 .
19 )

1

P ( z

1 .
19 )

1

.
8830

.
117
Thus:
P(x < 30 minutes) = .117
-1.19
0 1.19
z

(c)
x
   z
.
10
 
49

1 .
285 ( 16 )

70 minutes or more