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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Normal Probability Distribution
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 Distribution
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.
Normal Probability Distribution
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 .
Normal Probability Distribution
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.
Normal Probability Distribution
Characteristics
x
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Normal Probability Distribution
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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Normal Probability Distribution
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).
Normal Probability Distribution
Characteristics
.5 .5
x
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Normal Probability Distribution
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
of values of a normal random variable are within of its mean. of values of a normal random variable are within of its mean.95.44%95.44%
+/- 2 standard deviations+/- 2 standard deviations
of values of a normal random variable are within of its mean. of values of a normal random variable are within of its mean.99.72%99.72%
+/- 3 standard deviations+/- 3 standard deviations
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Normal Probability Distribution
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.
Standard Normal Probability Distribution
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Table of Cumulative Distribution FunctionTable of Cumulative Distribution Function
Standard Normal Probability Distribution
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
Standard Normal Probability Distribution
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Exercise 1Exercise 1
Standard Normal Probability Distribution
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)
Standard Normal Probability Distribution
Answer: P(z ≤ 1.00) = 0.8413 Answer: P(z ≤ 1.00) = 0.8413
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Exercise 2Exercise 2
Standard Normal Probability Distribution
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)
Standard Normal Probability Distribution
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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Exercise 3Exercise 3
Standard Normal Probability Distribution
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)
Standard Normal Probability Distribution
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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Exercise 4Exercise 4
Standard Normal Probability Distribution
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)
Standard Normal Probability Distribution
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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Exercise 5Exercise 5
Standard Normal Probability Distribution
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
Standard Normal Probability Distribution
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
Standard Normal Probability Distribution
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
Standard Normal Probability Distribution
zx
We can think of z as a measure of the number ofstandard deviations x is from .
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Exercise 6Exercise 6
Standard Normal Probability Distribution
Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?
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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Exercise 6Exercise 6
Standard Normal Probability Distribution
Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?
Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?
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
Standard Normal Probability Distribution
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Exercise 1Exercise 1
Standard Normal Probability Distribution
Find P(z ≤ 1.00)Find P(z ≤ 1.00)
Ans:=NORMSDIST(1)=0.8413
Ans:=NORMSDIST(1)=0.8413
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Exercise 2Exercise 2
Standard Normal Probability Distribution
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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Exercise 3Exercise 3
Standard Normal Probability Distribution
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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Exercise 4Exercise 4
Standard Normal Probability Distribution
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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Exercise 5Exercise 5
Standard Normal Probability Distribution
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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Exercise 6Exercise 6
Standard Normal Probability Distribution
Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?
Suppose the population mean is 10 and populationstandard deviation is 2. What is the probability thatthe random variable x is between 10 and 14?
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
Using Excel to Compute AdditionalStandard Normal Probabilities
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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Excel Formula Worksheet
Using Excel to Compute AdditionalStandard Normal Probabilities
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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Excel Value Worksheet
Using Excel to Compute AdditionalStandard Normal Probabilities
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 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.
Standard Normal Probability Distribution
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
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
Standard Normal Probability Distribution
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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)
Standard Normal Probability Distribution
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P(z > .83) = 1 – P(z < .83) = 1- .7967
= .2033
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.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)
Standard Normal Probability Distribution
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Solving for the Stockout Probability
0 .83
Area = .7967Area = 1 - .7967
= .2033
z
PepZone5w-20
Motor Oil
Standard Normal Probability Distribution
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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 thecorresponding z value.)
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Solving for the Reorder Point
PepZone5w-20
Motor Oil
0
Area = .9500
Area = .0500
zz.05
Standard Normal Probability Distribution
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Solving for the Reorder Point
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)
Standard Normal Probability Distribution
Excel Approach: z = NORMSINV(0.95) = 1.645
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Solving for the Reorder Point
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.
Standard Normal Probability Distribution
𝒛=𝒙−𝝁𝝈
conversion function
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Solving for the Reorder Point
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.
Standard Normal Probability Distribution
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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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Using Excel to ComputeNormal Probabilities
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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Excel Formula Worksheet
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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Excel Value Worksheet
Using Excel to ComputeNormal Probabilities
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