stat chapter 6 b
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8/10/2019 Stat Chapter 6 b
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Converting to a Standard Normal Distribution
x
z
Think of me as the measure
of the distance from the
mean, measured in
standard deviations
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is used to compute the zvalue
given a cumulative probability.
NORMSINV
NORM S INV
is used to compute the cumulative
probability given a zvalue.
NORMSDIST
NORM S DIST
Using Excel to ComputeStandard Normal Probabilities
Excel has two functions for computing probabilities
and zvalues for a standard normal distribution:
(The S in the function names remindsus that they relate to the standardnormal probability distribution.)
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Formula Worksheet
Using Excel to ComputeStandard Normal Probabilities
A B
1
2
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) =1-NORMSDIST(1.58)
8 P(z< -0.50) =NORMSDIST(-0.5)
9
Probabilities: Standard Normal Distribution
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Value Worksheet
Using Excel to ComputeStandard Normal Probabilities
A B
1
2
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) 0.0571
8 P(z< -0.50) 0.3085
9
Probabilities: Standard Normal Distribution
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Formula Worksheet
Using Excel to ComputeStandard Normal Probabilities
A B
1
2
3 zvalue with .10 in upper tail =NORMSINV(0.9)
4 zvalue with .025 in upper tail =NORMSINV(0.975)5 zvalue with .025 in lower tail =NORMSINV(0.025)
6
Finding zValues, Given Probabilities
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Value Worksheet
A B
1
2
3 zvalue with .10 in upper tail 1.28
4 zvalue with .025 in upper tail 1.965 zvalue with .025 in lower tail -1.96
6
Finding zValues, Given Probabilities
Using Excel to ComputeStandard Normal Probabilities
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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.
PepZone
5w-20Motor Oil
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Example: Pep Zone
Standard Normal Probability Distribution
The store manager is concerned that sales arebeing lost due to stockouts while waiting for anorder. It has been determined that demand duringreplenishment lead time is normally distributed with
a mean of 15 gallons and a standard deviation of 6gallons.
The manager would like to know the probabilityof a stockout, P(x> 20).
PepZone
5w-20Motor Oil
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Solving for Stockout Probability PepZone5w-20
Motor Oil
Step 1:Convert x to the standard normal distribution
83.6
1520
xz
Thus 20 gallons sold during the
replenishment lead time would be
.83 standard deviations above the
average of 15.
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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
. . . . . . . . . . .
Example: Pep ZonePep
Zone5w-20
Motor Oil
P(z .83) = 1 P(z< .83)
= 1- .7967
= .2033
Solving for the Stockout Probability
Example: Pep Zone
Step 3: Compute the area under the standard normalcurve to the right of z= .83.
PepZone
5w-20Motor Oil
Probabilityof a stockout P(x> 20)
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Solving for the Stockout Probability
Example: Pep Zone
0 .83
Area = .7967Area = 1 - .7967
= .2033
z
PepZone
5w-20Motor Oil
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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 PepZone
5w-20Motor Oil
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Solving for the Reorder Point
Example: Pep ZonePep
Zone5w-20
Motor Oil
0
Area = .9500
Area = .0500
z
z.05
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Solving for the Reorder Point
Example: Pep ZonePep
Zone5w-20
Motor Oil
Step 1: Find the z-value that cuts off an area of .05in the right tail of the standard normaldistribution.
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 .9738 .9744 .9750 .9756 .9761 .9767
. . . . . . . . . . .
We look up the complementof the tail area (1 - .05 = .95)
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Solving for the Reorder Point
Step 2: Convert z.05to the corresponding value ofx:
87.24)6(645.11505. zx
PepZone
5w-20Motor Oil
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Solving for the Reorder Point
So if we raising our reorder point
from 20 to 25 gallons, we reducethe probability of a stockout from
about .20 to less than .05
PepZone
5w-20Motor Oil
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Using Excel to ComputeNormal Probabilities
Excel has two functions for computing cumulative
probabilities and xvalues for any normaldistribution:
NORMDIST is used to compute the cumulativeprobability given an xvalue.
NORMINV is used to compute the xvalue givena cumulative probability.
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Formula Worksheet
Using Excel to ComputeNormal Probabilities
A B
1
2
3 P(x> 20) =1-NORMDIST(20,15,6,TRUE)
4 5
6
7 xvalue with .05 in upper tail =NORMINV(0.95,15,6)
8
Probabilities: Normal Distribution
Finding xValues, Given Probabilities
PepZone
5w-20Motor Oil
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Value Worksheet
Using Excel to ComputeNormal Probabilities
Note: P(x> 20) = .2023 here using Excel, while ourprevious manual approach using the ztable yielded.2033 due to our rounding of the zvalue.
A B
1
2
3 P(x> 20) 0.2023
4 5
6
7 xvalue with .05 in upper tail 24.87
8
Probabilities: Normal Distribution
Finding xValues, Given Probabilities
PepZone
5w-20Motor Oil
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Exercise 18, p. 261
The average time a subscriber reads the Wall Street Journalis 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 theJournal?
b) What is the probability a reader will spend no more than 30minutes reading the Journal?
c) For the 10 percent who spend the most time reading the
Journal, how much time do they spend?
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Exercise 18, p. 261
6875.16
4960
xz
a) Convertx to the standard normal distribution:
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
19.116
4930
z
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P(x 30 minutes)
0z
1.19-1.19
Red-shaded
area is equal to
blue shaded
area
117.8830.1
)19.1(1)19.1(
zPzP
Thus:
P(x < 30 minutes) = .117
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Exercise 18, p. 261
moreorminutes70)16(285.149
10.
zx
(c)
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