1 1 Slide Continuous Probability Distributions n A continuous random variable can assume any value...

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Continuous Probability Distributions

A continuous random variable can assume any value in an interval on the real line or in a collection of intervals.

It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval.

x

f (x) Normal

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Continuous Probability Distributions

The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2.

x

f (x) Normal

x1 x1 x2 x2

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Normal Probability Distribution

The normal probability distribution is the most important distribution for describing a continuous random variable.

It is widely used in statistical inference.

x

f (x) Normal

x1 x1 x2 x2

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It has been used in a wide variety of applications:

Heightsof peopleHeights

of people

Scientific measurements

Scientific measurements

Test scoresTest

scores

Amountsof rainfall

Amountsof rainfall

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Normal Probability Density Function

2 2( ) / 21( )

2xf x e

= mean = standard deviation = 3.14159e = 2.71828

where:

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The distribution is symmetric; its skewness measure is zero. The distribution is symmetric; its skewness measure is zero.


Characteristics

x

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The entire family of normal probability distributions is defined by its mean m and its standard deviation s .

The entire family of normal probability distributions is defined by its mean m and its standard deviation s .


Characteristics

Standard Deviation s

Mean mx

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The highest point on the normal curve is at the mean, which is also the median and mode. The highest point on the normal curve is at the mean, which is also the median and mode.


Characteristics

x

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Characteristics

-10 0 20

The mean can be any numerical value: negative, zero, or positive. The mean can be any numerical value: negative, zero, or positive.

x

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Characteristics

s = 15

s = 25

The standard deviation determines the width of thecurve: larger values result in wider, flatter curves.The standard deviation determines the width of thecurve: larger values result in wider, flatter curves.

x

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Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and .5 to the right).

Probabilities for the normal random variable are given by areas under the curve. The total area under the curve is 1 (.5 to the left of the mean and .5 to the right).


Characteristics

.5 .5

x

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Characteristics

of values of a normal random variable are within of its mean. of values of a normal random variable are within of its mean.68.26%68.26%

+/- 1 standard deviation+/- 1 standard deviation


+/- 2 standard deviations+/- 2 standard deviations


+/- 3 standard deviations+/- 3 standard deviations

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Characteristics

xm – 3s m – 1s

m – 2sm + 1s

m + 2sm + 3sm

68.26%

95.44%99.72%

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Standard Normal Probability Distribution

A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution.

A random variable having a normal distribution with a mean of 0 and a standard deviation of 1 is said to have a standard normal probability distribution.

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s = 1

0z

The letter z is used to designate the standard normal random variable. The letter z is used to designate the standard normal random variable.


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Table of Cumulative Distribution FunctionTable of Cumulative Distribution Function


Traditional way calculating probability withoutusing computer programs.Traditional way calculating probability withoutusing computer programs.

Make sure you understand the meaning of the tablebefore using it.Make sure you understand the meaning of the tablebefore using it.

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z Table is located in the Tables section of Contentson the Course Websitez Table is located in the Tables section of Contentson the Course Website


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Exercise 1Exercise 1


Find P(z ≤ 1.00)Find P(z ≤ 1.00)

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Exercise 1: Find P(z ≤ 1.00)Exercise 1: Find P(z ≤ 1.00)


Answer: P(z ≤ 1.00) = 0.8413 Answer: P(z ≤ 1.00) = 0.8413

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Find P(-0.5 ≤ z ≤ 1.25)Find P(-0.5 ≤ z ≤ 1.25)

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Exercise 2: Find P(-0.5 ≤ z ≤ 1.25)Exercise 2: Find P(-0.5 ≤ z ≤ 1.25)


Answer: P(-0.5 ≤ z ≤ 1.25) = P(z ≤ 1.25) - P(z ≤ -0.5)=Area I – Area II=0.8944 – 0.3085 = 0.5859

Answer: P(-0.5 ≤ z ≤ 1.25) = P(z ≤ 1.25) - P(z ≤ -0.5)=Area I – Area II=0.8944 – 0.3085 = 0.5859

= -

Area I Area II

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Find P(-1 ≤ z ≤ 1)Find P(-1 ≤ z ≤ 1)

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Exercise 3: Find P(-1 ≤ z ≤ 1)Exercise 3: Find P(-1 ≤ z ≤ 1)


Answer: P(-1 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ -1)=Area I – Area II=0.8413 – 0.1587= 0.6826

Answer: P(-1 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ -1)=Area I – Area II=0.8413 – 0.1587= 0.6826

= -

Area I Area II

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Find P(z ≥ 1.58)Find P(z ≥ 1.58)

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Exercise 4: Find P(z ≥ 1.58)Exercise 4: Find P(z ≥ 1.58)


Answer: P(z ≥ 1.58) = 1 - P(z ≤ 1.58) = 1 – 0.9429=0.0571Answer: P(z ≥ 1.58) = 1 - P(z ≤ 1.58) = 1 – 0.9429=0.0571

= -

Area I Area II

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Find z for cumulative probability = 0.1 in the uppertailFind z for cumulative probability = 0.1 in the uppertail

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Exercise 5: Find z for cumulative probability = 0.1in the upper tailExercise 5: Find z for cumulative probability = 0.1in the upper tail


Probability is 0.1in the upper tail

What is the z value here?

The lower-tail probability is1 - 0.1 = 0.9

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Exercise 5: Find z for cumulative probability = 0.1in the upper tailExercise 5: Find z for cumulative probability = 0.1in the upper tail


Answer:1. Look up the cumulative probability value closestto 0.9 (1-0.1) 0.88972. Use the z value found for the estimation 1.28

Answer:1. Look up the cumulative probability value closestto 0.9 (1-0.1) 0.88972. Use the z value found for the estimation 1.28

Probability valuesFind the closest one

to 0.9

0.9 is here

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Converting to the Standard Normal Distribution


zx

We can think of z as a measure of the number ofstandard deviations x is from .

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Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?


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Answer: When x=10, z=(10-10)/2 = 0When x=14, z=(14-10)/2 = 2Thus, P(0 ≤ z ≤ 2) = P(z ≤ 2) - P(z ≤ 0)=0.9772 – 0.5000 = 0.4772

Answer: When x=10, z=(10-10)/2 = 0When x=14, z=(14-10)/2 = 2Thus, P(0 ≤ z ≤ 2) = P(z ≤ 2) - P(z ≤ 0)=0.9772 – 0.5000 = 0.4772

zx

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is used to compute the z value given a LEFT-TAIL cumulative probability. is used to compute the z value given a LEFT-TAIL cumulative probability.NORMSINVNORMSINVNORM S INV

is used to compute the LEFT-TAILcumulative probability given a z value. is used to compute the LEFT-TAILcumulative probability given a z value.NORMSDISTNORMSDISTNORM S DIST

Using Excel to ComputeStandard Normal Probabilities

Excel has two functions for computing probabilities and z values for a standard normal distribution:

(The “S” in the function names remindsus that they relate to the standardnormal probability distribution.)

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Practice Exercise 1 to 6 by using ExcelPractice Exercise 1 to 6 by using Excel


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Find P(z ≤ 1.00)Find P(z ≤ 1.00)

Ans:=NORMSDIST(1)=0.8413

Ans:=NORMSDIST(1)=0.8413

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Find P(-0.5 ≤ z ≤ 1.25)Find P(-0.5 ≤ z ≤ 1.25)

Answer: P(-0.5 ≤ z ≤ 1.25) = P(z ≤ 1.25) - P(z ≤ -0.5)=Area I – Area II=NORMSDIST(1.25)-NORMSDIST(-0.5)=0.8944 – 0.3085 = 0.5858

Answer: P(-0.5 ≤ z ≤ 1.25) = P(z ≤ 1.25) - P(z ≤ -0.5)=Area I – Area II=NORMSDIST(1.25)-NORMSDIST(-0.5)=0.8944 – 0.3085 = 0.5858

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Find P(-1 ≤ z ≤ 1)Find P(-1 ≤ z ≤ 1)

Answer: P(-1 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ -1)=Area I – Area II=NORMSDIST(1)-NORMSDIST(-1)=0.8413 – 0.1587= 0.6827

Answer: P(-1 ≤ z ≤ 1) = P(z ≤ 1) - P(z ≤ -1)=Area I – Area II=NORMSDIST(1)-NORMSDIST(-1)=0.8413 – 0.1587= 0.6827

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Find P(z ≥ 1.58)Find P(z ≥ 1.58)

Answer: P(z ≥ 1.58) = 1 - P(z ≤ 1.58) = 1 – NORMSDIST(1.58)= 1 – 0.9429=0.0571

Answer: P(z ≥ 1.58) = 1 - P(z ≤ 1.58) = 1 – NORMSDIST(1.58)= 1 – 0.9429=0.0571

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Find z for cumulative probability = 0.1 in the uppertailFind z for cumulative probability = 0.1 in the uppertail

Answer:= NORMSINV(1-0.1) or =NORMSINV(0.9)=1.28

Answer:= NORMSINV(1-0.1) or =NORMSINV(0.9)=1.28

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Answer: When x=10, z=(10-10)/2 = 0When x=14, z=(14-10)/2 = 2Thus, P(0 ≤ z ≤ 2) = P(z ≤ 2) - P(z ≤ 0)=NORMSDIST(2)-NORMSDIST(0)=0.9773 – 0.5000 = 0.4773

Answer: When x=10, z=(10-10)/2 = 0When x=14, z=(14-10)/2 = 2Thus, P(0 ≤ z ≤ 2) = P(z ≤ 2) - P(z ≤ 0)=NORMSDIST(2)-NORMSDIST(0)=0.9773 – 0.5000 = 0.4773

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Excel Formula View

Using Excel to ComputeStandard Normal Probabilities

Excel Worksheet View

See 6Exercises-key.xlsx

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Excel Formula Worksheet

Using Excel to Compute AdditionalStandard Normal Probabilities

A B12 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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Excel Value Worksheet


A B12 3 P (z < 1.00) 0.84134 P (0.00 < z < 1.00) 0.34135 P (0.00 < z < 1.25) 0.39446 P (-1.00 < z < 1.00) 0.68277 P (z > 1.58) 0.05718 P (z < -0.50) 0.30859

Probabilities: Standard Normal Distribution

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A B

12 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

Finding z Values, Given Probabilities

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A B

12 3 z value with .10 in upper tail 1.284 z value with .025 in upper tail 1.965 z value with .025 in lower tail -1.966

Finding z Values, Given Probabilities

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Example: Pep Zone

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 areplenishment order.

PepZone5w-20Motor Oil

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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.


Example: Pep Zone

The manager would like to know the probability

of a stockout during replenishment lead-time. In

other words, what is the probability that demand

during lead-time will exceed 20 gallons? P(x > 20) = ?

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Example: Pep Zone

𝜇=15

𝜎=6

𝑥=20

𝑯𝒐𝒘𝒃𝒊𝒈 𝒊𝒔𝒕𝒉𝒆𝒂𝒓𝒆𝒂?

𝑥

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z = (x - )/ = (20 - 15)/6 = .83

z = (x - )/ = (20 - 15)/6 = .83

Solving for the Stockout Probability

Step 1: Convert x to the standard normal distribution.Step 1: Convert x to the standard normal distribution.

PepZone5w-20

Motor Oil

Step 2: Find the area under the standard normal curve to the left of z = .83.Step 2: Find the area under the standard normal curve to the left of z = .83.

see next slide see next slide


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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

. . . . . . . . . . .

PepZone5w-20

Motor Oil

P(z < .83)


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P(z > .83) = 1 – P(z < .83) = 1- .7967

= .2033

P(z > .83) = 1 – P(z < .83) = 1- .7967

= .2033


Step 3: Compute the area under the standard normal curve to the right of z = .83.Step 3: Compute the area under the standard normal curve to the right of z = .83.

PepZone5w-20

Motor Oil

Probability of a

stockoutP(x > 20)


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0 .83

Area = .7967Area = 1 - .7967

= .2033

z

PepZone5w-20

Motor Oil


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If the manager of Pep Zone wants the probability

of a stockout during replenishment lead-time to be

no more than .05, what should the reorder point be?

---------------------------------------------------------------

(Hint: Given a probability, we can use the standard

normal table in an inverse fashion to find thecorresponding z value.)

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Solving for the Reorder Point

PepZone5w-20

Motor Oil

0

Area = .9500

Area = .0500

zz.05


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PepZone5w-20

Motor Oil

Step 1: Find the z-value that cuts off an area of .05 in the right tail of the standard normal distribution.

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 .9738 .9744 .9750 .9756 .9761 .9767

. . . . . . . . . . .We look up the

complement of the tail area (1 - .05 = .95)


Excel Approach: z = NORMSINV(0.95) = 1.645

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PepZone5w-20

Motor Oil

Step 2: Convert z.05 to the corresponding value of x.Step 2: Convert z.05 to the corresponding value of x.

x = + z.05 = 15 + 1.645(6) = 24.87 or 25

x = + z.05 = 15 + 1.645(6) = 24.87 or 25

A reorder point of 25 gallons will place the probability of a stockout during leadtime at (slightly less than) .05.


𝒛=𝒙−𝝁𝝈

conversion function

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PepZone5w-20

Motor Oil

By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockoutdecreases from about .20 to .05. This is a significant decrease in the chance that PepZone will be out of stock and unable to meet acustomer’s desire to make a purchase.


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Using Excel to ComputeNormal Probabilities

Excel has two functions for computing cumulative probabilities and x values for any normal distribution:

NORMDIST is used to compute the LEFT-TAIL cumulative probability given an x value.NORMDIST is used to compute the LEFT-TAIL cumulative probability given an x value.

NORMINV is used to compute the x value givena LEFT-TAIL cumulative probability.NORMINV is used to compute the x value givena LEFT-TAIL cumulative probability.

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Syntax: NORMDIST(x, mu, sigma, cumulative)

mu: population mean sigma: population standard deviation cumulative:

true—probability is returned false – height of the bell-shaped curve is

returned NORMINV(p, mu, sigma)

p: probability

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Using Excel to ComputeNormal Probabilities Pep

Zone5w-20

Motor Oil

A B

12 3 P (x > 20) =1-NORMDIST(20,15,6,TRUE)4 56 7 x value with .05 in upper tail =NORMINV(0.95,15,6)8

Probabilities: Normal Distribution

Finding x Values, Given Probabilities

Lower tail probability: NORMDIST(20, 15, 6, TRUE)Upper tail probability: 1-NORMDIST(20, 15, 6, TRUE)

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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.

PepZone5w-20

Motor Oil

A B

12 3 P (x > 20) 0.20234 56 7 x value with .05 in upper tail 24.878

Probabilities: Normal Distribution

Finding x Values, Given Probabilities

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