Stat Chapter 6 b

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


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


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


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



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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


Zone5w-20

Motor Oil

0

Area = .9500

Area = .0500

z

z.05


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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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Step 2: Convert z.05to the corresponding value ofx:

87.24)6(645.11505. zx

PepZone

5w-20Motor Oil


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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


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


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)

Stat Chapter 6 b

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