M11-Normal Distribution 1 1 Department of ISM, University of Alabama, 1995-2003
Lesson Objective
Learn the mechanics of using the table for the Normal Distribution.
Given a region for a variable that follows the Normal Distribution,find the probability that a randomlyselected item will fall in this region.
Given a probability, find the region for a normally distributed variable that corresponds to this probability.
M12-Normal Distribution 2 2 Department of ISM, University of Alabama, 1995-2003
The The NormalNormal DistributionsDistributions
a.k.a., “a.k.a., “The The Bell Shaped CurveBell Shaped Curve””
Describes the shape for some quantitative, continuous random variables.
M12-Normal Distribution 2 3 Department of ISM, University of Alabama, 1995-2003
= mean determines the location.
= standard deviation determines spread, variation, scatter.
Normal Population Distribution Normal Population Distribution has two has two parametersparameters::
M12-Normal Distribution 2 4 Department of ISM, University of Alabama, 1995-2003
X ~ N( = 66, = 9) or N(66, 9)
Z = the number of standard deviations that an X - value is from the mean.
Notation:Notation:
X - Z =
Z ~ N( = 0, = 1 ) or N(0,1)
Z follows the “Standard Normal Distribution”
M12-Normal Distribution 2 5 Department of ISM, University of Alabama, 1995-2003
-4 -3 -2 -1 0 1 2 3 4
____, ±1
______, ±2
______, ±3
Empirical Rule of the Normal Distribution
Where does this
come from?
M12-Normal Distribution 2 6 Department of ISM, University of Alabama, 1995-2003
Recall
The “area” under the curve within a range of X values is equal to proportion of the population within that range of X values.
Question: How do we compute “areas”?
• Geometry formulas
• Calculus (integration)
• Tables
• Excel
• Minitab
M12-Normal Distribution 2 7 Department of ISM, University of Alabama, 1995-2003
Reading the Standard Normal Table(finding areas under the normal curve)
Step 1 for all problems:
DTDPDTDP
M12-Normal Distribution 2 8 Department of ISM, University of Alabama, 1995-2003
0 1.72
Table gives P(0 < z < ?) =Table gives
P(0 < z < ?) =
FindP(0 < z < 1.72) =
Up to the 1st decimal placeUp to the 1st decimal place
2nd decimal place2nd decimal place
.4573.4573
Standard Standard Normal Normal
TableTable
Standard Standard Normal Normal
TableTable
M12 Normal Distribution 2 9 Department of ISM, University of Alabama, 1995-2003
P(-1.23 < Z < 2.05) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z2.05
What proportion of Z values are between –1.23 and +2.05?
P(1.23 < Z < 2.05) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z
What proportion of Z values are between +1.23 and +2.05?
= =
=
= 1.23
-1.23
2.05
?
M12 Normal Distribution 2 10 Department of ISM, University of Alabama, 1995-2003
P( 10.0 < X < 15.72) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
Z
P( 0 < Z < 1.43)
Weights of packages are normally
distributed with mean of 10 lbs. and standard deviation of 4.0 lbs. Find the proportion of packages that weigh between 10 and 15.72 lbs.
10 X15.72
Z = 15.72 – 10.0 4.0
=
Z = 10.0 – 10.0 4.0
=
X = weight of packages.
X ~ N( = 10, = 4.0)
=
M12 Normal Distribution 2 11 Department of ISM, University of Alabama, 1995-2003
P( X > 15.72) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z1.43
P( Z > 1.43)
Same situation.
What proportion of packages weigh more than 15.72 lbs?
10 X15.72
Z = 15.72 – 10.0 4.0
= 1.43
X = weight of packages.
X ~ N( = 10, = 4.0)
==
?
M12 Normal Distribution 2 12 Department of ISM, University of Alabama, 1995-2003
P( X < 14.2) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z1.05
.3531
P( Z < 1.05)
Same situation.
What proportion of packages weigh less than 14.2 lbs?
10 X14.2
Z = 14.2 – 10.0 4.0
= 1.05
X = weight of packages.
X ~ N( = 10, = 4.0)
= .5 + .3531= .8531
.5000
M12-Normal Distribution 2 13 Department of ISM, University of Alabama, 1995-2003
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
Same situation.
What proportion of packages weigh between 5.08 and 18.2 lbs?
X = weight of packages.
X ~ N( = 10, = 4.0)
M12-Normal Distribution 2 14 Department of ISM, University of Alabama, 1995-2003
Same situation.
What proportion of packages weigh either less than 2.4 lbs or greater than 11.0 lbs?
X = weight of packages.
X ~ N( = 10, = 4.0)
Homework
M12 Normal Distribution 2 15 Department of ISM, University of Alabama, 1995-2003
P( X < ?) = .10
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z-1.28
.10
P( Z < )
Same situation.
Find the weight such that 10% of all packages weigh less than this weight.
10 X?
X = weight of packages.
X ~ N( = 10, = 4.0)
= .10
.40
–1.28 =? – 10 4
? = 10 – 1.28 • 4 = 10 – 5.12 = 4.88 pounds
-1.28
This is a backwards problem! We are given the probability;
we need to find the boundary.
10% weigh less than 4.88 pounds;90% weigh more than 4.88 pounds.10% weigh less than 4.88 pounds;90% weigh more than 4.88 pounds.
M12 Normal Distribution 2 16 Department of ISM, University of Alabama, 1995-2003
0
Table gives P(0 < z < ?) =Table gives
P(0 < z < ?) =
Standard Standard Normal Normal
TableTable
Standard Standard Normal Normal
TableTable
FindP( __ < z < 0) =.40
.4000.4000
-1.28
Find the Z value to cut off the top 10%.
M12 Normal Distribution 2 17 Department of ISM, University of Alabama, 1995-2003
0
Table gives P(0 < z < ?) =Table gives
P(0 < z < ?) =
Standard Standard Normal Normal
TableTable
Standard Standard Normal Normal
TableTable
.25.25
?
Find the Z values that define the middle 50%.
.25.25
?
M12 Normal Distribution 2 18 Department of ISM, University of Alabama, 1995-2003
Table gives P(0 < z < ?) =Table gives
P(0 < z < ?) =
Standard Standard Normal Normal
TableTable
Standard Standard Normal Normal
TableTable
Find the Z values that define the middle 95%.
M12-Normal Distribution 2 19 Department of ISM, University of Alabama, 1995-2003
Normal Functions in Excel
• NORMDIST – Used to compute areas under any normal curve. Can also compute height of curve (not useful except for drawing normal curves).
• NORMSDIST - Used to compute areas under a standard normal( N(0,1) or Z curve ).
M12-Normal Distribution 2 20 Department of ISM, University of Alabama, 1995-2003
Normal Functions in Excel
• NORMINV - Used to find the X value corresponding to a given cumulative probability for any normal distribution.
• NORMSINV - Used to find the Z value corresponding to a given cumulative probability for a standard normal distribution.
M12-Normal Distribution 2 21 Department of ISM, University of Alabama, 1995-2003
1. P( Z < –1.92)
=2. P( Z < 2.56) =3. P( Z > 0.80) =4. P( Z = 1.42) =5. P( .32 < Z < 2.48) =6. P( -1.75 < Z < 1.75) =7. P( Z < 4.25) =
8. P( Z > 4.25) =
9. P(-.05 < Z < .05) =
10. Find Z such that only 12% are smaller.
Practice problems.You MUST know how to work ALL of these problems and the followingpractice problemsto pass this course.
M12-Normal Distribution 2 22 Department of ISM, University of Alabama, 1995-2003
1. .02742. .99483. 1.0 – .7881 = .21194. .05. .9934 – .6255 = .06896. .9599 – .0401 = .91987. 1.00008. .00009. .039810. -1.175
Practice problem answers
M12-Normal Distribution 2 23 Department of ISM, University of Alabama, 1995-2003
Question:What do we do when we have a normal population distribution, but the mean is not “0” and/or the standard deviation is not “1”?
Z = X –
Use the Universal Translator
Example: Suppose X ~ N(120, 10).
11. Find P ( X > 150 ).
12. Find the quartiles of this distribution.
M12-Normal Distribution 2 24 Department of ISM, University of Alabama, 1995-2003
P( 0 < Z < 1.43) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z1.43
What proportion of Z values are between 0 and 1.43?
.4236
= .4236
P(-1.43 < Z < 0) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z-1.43
What proportion of Z values are between -1.43 and 0?
.4236
= .4236
ori
M12-Normal Distribution 2 25 Department of ISM, University of Alabama, 1995-2003
P(Z > 1.43) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z1.43
What proportion of Z values are greater than 1.43?
.4236
P(Z < 1.43) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z
What proportion of Z values are less than 1.43?
.4236
.5000
.0764
= .5 - .4236 = .0764
= .5 + .4236 = .9236 1.43
.5000
M12-Normal Distribution 2 26 Department of ISM, University of Alabama, 1995-2003
P(-1.23 < Z < 2.05) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z2.05
What proportion of Z values are between –1.23 and 2.05?
.3907
P(1.23 < Z < 2.05) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z
What proportion of Z values are between 1.23 and 2.05?
.4236
.4798
= .4798 + .3907 = .8705
= .4798 - .4236 = .0562 1.23
.4798
-1.23
2.05
.0562
M12-Normal Distribution 2 27 Department of ISM, University of Alabama, 1995-2003
P( 4.28 < X < 10.0) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z-1.43
.4236
= P( -1.43 < Z < 0)
Same situation.
Find the proportion of packages that weigh between 4.28 and 10.0 lbs.
10 X4.28
Z = 4.28 – 10.0 4.0
= -1.43
Z = 10.0 – 10.0 4.0
= 0
X = weight of packages.
X ~ N( = 10, = 4.0)
= .4236
M12-Normal Distribution 2 28 Department of ISM, University of Alabama, 1995-2003
P( 13.0 < X < 17.84) = ?
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
0 Z1.96
.4750
= P( .75 < Z < 1.96)
Same situation.
What proportion of the packages weigh between 13.0 and 17.84 lbs?
10 X17.84
Z = 17.84 – 10.0 4.0
= 1.96
X = weight of packages.
X ~ N( = 10, = 4.0)
= .4750 - .2734= .2016
.2734
.7513
Z = 13.0 – 10.0 4.0
= .75
?
M12-Normal Distribution 2 29 Department of ISM, University of Alabama, 1995-2003
-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0
Z
Same situation.
Find the weight such that a. 16% weigh more less than this value.b. You have the boundaries of the middle 80%.c. The top 25% weigh more.d. You have the quartiles.
X
X = weight of packages.
X ~ N( = 10, = 4.0)
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