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Unformatted text preview: Chapter 6
Continuous Probability
Distributions • Uniform Probability Distribution
• Normal Probability Distribution
• Exponential Probability Distribution
f (x) f (x) Uniform
f (x) Normal
x x
x Exponential 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. Continuous Probability Distributions
s 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.
Uniform f (x) f (x) x x
1 Normal x x x x x 1 2 x x
1 Exponential f (x) 2 x 12 2 x Uniform Probability Distribution
• A random variable is uniformly distributed whenever the probability is proportional to the interval’s length. • The uniform probability density function is: f (x) = 1/(b – a) for a < x < b = 0 elsewhere
where: a = smallest value the variable can assume b = largest value the variable can assume Uniform Probability Distribution
• Expected Value of x
E(x) = (a + b)/2 • Variance of x
Var(x) = (b a)2/12 Uniform Probability Distribution
• Example: Slater's Buffet
Slater customers are charged
for the amount of salad they take. Sampling suggests that the
amount of salad taken is uniformly distributed
between 5 ounces and 15 ounces. Uniform Probability
Distribution
s Uniform Probability Density Function f(x) = 1/10 for 5 < x < 15 = 0 elsewhere
where: x = salad plate filling weight Uniform Probability
Distribution
s Expected Value of x E(x) = (a + b)/2 = (5 + 15)/2 = 10 s Variance of x Var(x) = (b a)2/12 = (15 – 5)2/12 = 8.33 Uniform Probability Distribution
• Uniform Probability Distribution
for Salad Plate Filling Weight
f( x ) 1/10
5
10
15
Salad Weight (oz.) x Uniform Probability
Distribution What is the probability that a customer will take between 12 and 15 ounces of salad?
f( x )
P(12 < x < 15) = 1/10(3) = .3
1/10
5
10 12
15
Salad Weight (oz.) x 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. Normal Probability Distribution
s It has been used in a wide variety of applications:
Heights
of people Scientific measurements Normal Probability Distribution
s It has been used in a wide variety of applications:
Test scores Amounts
of rainfall Normal Probability Distribution
• Normal Probability Density Function 1
− ( x − µ )2 / 2 σ 2
f (x ) =
e
σ 2π
where: µ = mean
σ = standard deviation
π = 3.14159
e = 2.71828 Normal Probability Distribution
s Characteristics The distribution is symmetric; its skewness measure is zero. x Normal Probability Distribution
s Characteristics The entire family of normal probability distributions is defined by its mean µ and its standard deviation σ .
Standard Deviation σ Mean µ x Normal Probability Distribution
s Characteristics The highest point on the normal curve is at the mean, which is also the median and mode. x Normal Probability Distribution
s Characteristics The mean can be any numerical value: negative, zero, or positive. 10 0 20 x Normal Probability Distribution
s Characteristics
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves. σ = 15 σ = 25
x Normal Probability Distribution
s Characteristics 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). .5 .5
x Normal Probability Distribution
s Characteristics
68.26% of values of a normal random variable
+/ 1 standard deviation are within of its mean.
95.44% of values of a normal random variable
+/ 2 standard deviations are within of its mean.
99.72% of values of a normal random variable
+/ 3 standard deviations are within of its mean. Normal Probability Distribution
s Characteristics
99.72%
95.44%
68.26% µ µ – 3σ
µ – 1σ
µ – 2σ µ + 3σ
µ + 1σ
µ + 2σ x 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. Standard Normal Probability Distribution The letter z is used to designate the standard normal random variable.
σ= 1 0 z Standard Normal Probability Distribution
s Converting to the Standard Normal Distribution x−µ
z=
σ
We can think of z as a measure of the number of
standard deviations x is from µ . Standard Normal Probability
Distribution
• Example: Pep Zone
Pep Zone sells auto parts and supplies including a popular multigrade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed.
Pep
Zone
5w20 Motor Oil Standard Normal Probability
Distribution
• Example: Pep Zone 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 leadtime is normally
Pep
distributed with a mean of 15 gallons and
Zone
5w20
a standard deviation of 6 gallons. Motor Oil
The manager would like to know the
probability of a stockout, P(x > 20). Standard Normal Probability
Distribution
s Solving for the Stockout Probability Step 1: Convert x to the standard normal distribution. z = (x µ )/σ = (20 15)/6 = .83
Step 2: Find the area under the standard normal curve to the left of z = .83. see next slide Pep
Zone
5w20
Motor Oil Standard Normal Probability
Distribution
Cumulative Probability Table for the Standard Normal Distribution s Pep
Zone
5w20
Motor Oil 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) . . . . . Standard Normal Probability
Distribution
s 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) Pep
Zone
5w20
Motor Oil Standard Normal Probability
Distribution
s Solving for the Stockout Probability Area = 1 .7967 Area = .7967 = .2033 0 .83 z Pep
Zone
5w20
Motor Oil Standard Normal Probability
Distribution Pep
Zone
5w20
Motor Oil • Standard Normal Probability Distribution
If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Standard Normal Probability
Distribution
s Solving for the Reorder Point Area = .9500
Area = .0500 z.05 0 z Pep
Zone
5w20
Motor Oil Standard Normal Probability
Distribution
s Solving for the Reorder Point Pep
Zone
5w20
Motor Oil Step 1: Find the zvalue 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) .
.
.
.
.
.
.
.
.
. Standard Normal Probability
Distribution
s Solving for the Reorder Point Step 2: Convert z.05 to the corresponding value of x. x = µ + z.05σ
= 15 + 1.645(6)
= 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. Pep
Zone
5w20
Motor Oil Standard Normal Probability
Distribution
s Solving for the Reorder Point By raising the reorder point from 20 gallons to 25 gallons on hand, the probability of a stockout
decreases from about .20 to .05. This is a significant decrease in the chance that Pep
Zone will be out of stock and unable to meet a
customer’s desire to make a purchase. Pep
Zone
5w20
Motor Oil Exponential Probability Distribution
s s The exponential probability distribution is useful in describing the time it takes to complete a task.
The exponential random variables can be used to describe:
Time between
vehicle arrivals
at a toll booth Time required
to complete
a questionnaire Distance between
major defects
in a highway
SLOW Exponential Probability
Distribution
• Density Function
1 − x /µ
f ( x ) = µ > 0
e
for x > 0, µ
where: µ = mean e = 2.71828 Exponential Probability
Distribution
• Cumulative Probabilities
P ( x ≤ x0 ) = 1 − e − xo / µ
where: x0 = some specific value of x Exponential Probability
Distribution
• Example: Al’s FullService Pump The time between arrivals of cars
at Al’s fullservice gas pump follows
an exponential probability distribution
with a mean time between arrivals of 3 minutes. Al would like to know the
probability that the time between two successive arrivals will be 2 minutes or less. Exponential Probability
Distribution
f(x)
.4 P(x < 2) = 1 2.718282/3 = 1 .5134 = .4866 .3
.2
.1
x 1 2 3 4 5 6 7 8 9 10 Time Between Successive Arrivals (mins.) ...
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This note was uploaded on 08/28/2011 for the course BUS 300 taught by Professor White during the Spring '09 term at Rutgers.
 Spring '09
 White

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