bia2610 – statistical methods chapter 6 – continuous probability distributions
TRANSCRIPT
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BIA2610 – Statistical MethodsChapter 6 – Continuous Probability Distributions
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Chapter 6 Continuous Probability Distributions
Uniform Probability Distribution
x
f (x) Exponential
Normal Probability Distribution Exponential Probability Distribution
f (x)
x
Uniform
x
f (x) Normal
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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.
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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.
f (x)
x
Uniform
x2 x1
x
f (x) Exponential
x1 x2
x
f (x)Normal
x1 x2
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Normal Probability Distribution
It is widely used in statistical inference. It has been used in a wide variety of applications
including:
• Heights of people• Rainfall amounts
• Test scores• Scientific measurements
The normal probability distribution is the most important distribution for describing a continuous random variable.
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Normal Probability Distribution
The distribution is symmetric; its skewness measure is zero.
Characteristics
x
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Normal Probability Distribution
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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Normal Probability Distribution
The highest point on the normal curve is at the mean, which is also the median and mode.
Characteristics
x
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Normal Probability Distribution
Characteristics
-10 0 25
The mean can be any numerical value: negative, zero, or positive.
x
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Normal Probability Distribution
Characteristics
s = 15
s = 25
The standard deviation determines the width of the curve: larger values result in wider, flatter curves.
x
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Normal Probability Distribution
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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Normal Probability Distribution
Characteristics (basis for the empirical rule)
of values of a normal random variable are within of its mean.
68.26%+/- 1 standard deviation
of values of a normal random variable are within of its mean.
95.44%+/- 2 standard deviations
of values of a normal random variable are within of its mean.
99.72%+/- 3 standard deviations
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Normal Probability Distribution
Characteristics (basis for the empirical rule)
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.
Characteristics
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Standard Normal Probability Distribution
s = 1
0z
The letter z is used to designate the standard normal random variable.
Characteristics
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Standard Normal Probability Distribution
Converting to the Standard Normal Distribution
We can think of z as a measure of the number ofstandard deviations x is from .
z =
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Using Excel to ComputeStandard Normal Probabilities
is used to compute the z value given a cumulative probability.
NORM.S.INV
is used to compute the cumulative probability given a z value.NORM.S.DIST
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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Using Excel to ComputeStandard Normal Probabilities
Excel Formula Worksheet
A B12 3 P (z < 1.00) =NORM.S.DIST(1)4 P (0.00 < z < 1.00) =NORM.S.DIST(1)-NORM.S.DIST(0)5 P (0.00 < z < 1.25) =NORM.S.DIST(1.25)-NORM.S.DIST(0)6 P (-1.00 < z < 1.00) =NORM.S.DIST(1)-NORM.S.DIST(-1)7 P (z > 1.58) =1-NORM.S.DIST(1.58)8 P (z < -0.50) =NORM.S.DIST(-0.5)9
Probabilities: Standard Normal Distribution
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Using Excel to ComputeStandard Normal Probabilities
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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Using Excel to ComputeStandard Normal Probabilities
Excel Formula Worksheet
A B
12 3 z value with .10 in upper tail =NORM.S.INV(0.9)4 z value with .025 in upper tail =NORM.S.INV(0.975)5 z value with .025 in lower tail =NORM.S.INV(0.025)6
Finding z Values, Given Probabilities
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Using Excel to ComputeStandard Normal Probabilities
Excel Value Worksheet
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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Standard Normal Probability Distribution
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 a
replenishment order.
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Standard Normal Probability Distribution
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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Standard Normal Probability Distribution
z = (x - )/ = (20 - 15)/6 = .83
Solving for the Stockout Probability
Step 1: Convert x to the standard normal distribution.
Step 2: Find the area under the standard normal curve to the left of z = .83.
see next slide
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Standard Normal Probability 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
. . . . . . . . . . .
Cumulative Probability Table for the Standard Normal Distribution
P(z < .83)
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Standard Normal Probability Distribution
P(z > .83) = 1 – P(z < .83) = 1- .7967
= .2033
Solving for the Stockout Probability
Step 3: Compute the area under the standard normal curve to the right of z = .83.
Probability of a stockout P(x > 20)
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Standard Normal Probability Distribution
Solving for the Stockout Probability
0 .83
Area = .7967Area = 1 - .7967
= .2033
z
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Standard Normal Probability Distribution
• Standard Normal Probability Distribution 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 the
corresponding z value.)
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Standard Normal Probability Distribution
Solving for the Reorder Point
0
Area = .9500
Area = .0500
zz.05
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Standard Normal Probability Distribution
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.
We look upthe complement of the tail area(1 - .05 = .95)
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 .97061.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
. . . . . . . . . . .
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Standard Normal Probability Distribution
Solving for the Reorder Point
Step 2: Convert z.05 to the corresponding value of x.
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.
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Standard Normal Probability Distribution
Solving for the Reorder Point
15x
24.87
Probability of astockout duringreplenishmentlead-time = .05
Probability of nostockout duringreplenishmentlead-time = .95
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Standard Normal Probability Distribution
Solving for the Reorder Point
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 thatPep Zone 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:
NORM.DIST is used to compute the cumulative probability given an x value.
NORM.INV is used to compute the x value given a cumulative probability.
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Using Excel to ComputeNormal Probabilities
Excel Formula Worksheet
A B
12 3 P (x > 20) =1-NORM.DIST(20,15,6,TRUE)4 56 7 x value with .05 in upper tail =NORM.INV(0.95,15,6)8
Probabilities: Normal Distribution
Finding x Values, Given Probabilities
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Using Excel to ComputeNormal Probabilities
Excel Formula Worksheet
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.
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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End of Chapter 6