Transcript Chapter 5

5.1 Normal Probability Distributions
Normal distribution
• A continuous probability distribution for a continuous random variable, x.
• The most important continuous probability distribution in statistics.
• The graph of a normal distribution is called the normal curve.
x
Larson/Farber
1
Properties of Normal Distributions
1.
2.
3.
4.
5.
The mean, median, and mode are equal.
The normal curve is bell-shaped and symmetric about the mean.
The total area under the curve is equal to one.
The normal curve approaches, but never touches the x-axis as it extends
farther and farther away from the mean.
Between μ – σ and μ + σ (in the center of the curve), the graph curves
downward. The graph curves upward to the left of μ – σ and to the right
of μ + σ. The points at which the curve changes from curving upward to
curving downward are called the inflection points.
Inflection points
Total area = 1
μ  3σ
μ  2σ
μσ
μ
μ+σ
μ + 2σ
μ + 3σ
x
2
Means and Standard Deviations
• A normal distribution can have any mean and any positive standard deviation.
• The mean gives the location of the line of symmetry.
• The standard deviation describes the spread of the data.
μ = 3.5
σ = 1.5
Larson/Farber
μ = 3.5
σ = 0.7
μ = 1.5
σ = 0.7
3
Example: Understanding Mean and
Standard Deviation
Which curve has the greater mean and Standard Deviation?
Curve A has the greater mean (The
line of symmetry of curve A occurs at x
= 15. The line of symmetry of curve B
occurs at x = 12.)
Curve B has the greater standard
deviation (Curve B is more spread
out than curve A.)
Larson/Farber
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Interpreting Graphs
The heights of fully grown white oak trees are normally
distributed. The curve represents the distribution.
σ = 3.5 (The inflection
points are one standard
deviation away from
the mean)
μ = 90 (A normal
curve is symmetric
about the mean)
Z-Scores:
Larson/Farber
3
-2
-1
0
1
2
3
5
The Standard Normal Distribution
Standard normal distribution
• A normal distribution with a mean of 0 and a standard
deviation of 1.
Area = 1
3
2
1
z
0
1
2
3
• Any x-value can be transformed into a z-score by
using the formula
Value - Mean
x-
z

Standard deviation

Larson/Farber
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The Standard Normal Distribution
• If each data value of a normally distributed random
variable x is transformed into a z-score, the result will
be the standard normal distribution.
Normal Distribution

z

x
x-
Standard Normal
Distribution

1
0
z
• Use Standard Normal Table or Calculator function: NormalCdf() to
find cumulative area under the standard normal curve.
Larson/Farber
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Properties of the Standard Normal
Distribution
1.
2.
3.
4.
The cumulative area is close to 0 for z-scores close to z = 3.49.
The cumulative area increases as the z-scores increase.
The cumulative area for z = 0 is 0.5000.
The cumulative area is close to 1 for z-scores close to z = 3.49.
** Note: Cumulative Area = Area under the curve to the ‘LEFT’ of the z-score.
Area is
close to 0
z = 3.49
Larson/Farber
Area is
close
to 1
z
3
2
1
0
Z=0
Area = 0.500
1
2
3
z = 3.49
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Using The Standard Normal Table
Find the cumulative area that corresponds to a z-score of 1.15 and –0.24
The area to the
left of z = 1.15 is
0.8749
TI-83/84
<2nd><DISTR>
2-NormalCdf
Normalcdf(-10000, 1.15)
The area to the
left of z = 0.24 is
0.4052
Larson/Farber
Why use –10000 in
normal cdf()? 9
Finding Areas Under the Standard
Normal Curve
#1 To find the area to the left of z, find #2 To find area to the right of z, find
the area that corresponds to z in
area corresponding to z in table &
the standard normal table or
subtract from 1 (total area) or
Ti83/84 normalcdf(-10000, zscore)
Ti83/84 normalcdf (zscore, 10000)
The area to the left
of z = 1.23 is 0.8907
Area to the left
of z = 1.23 is
0.8907.
Area to the right is
1 – 0.8907 = 0.1093
#3 To find the area between two z-scores, find both z-scores in table & subtract smaller
from larger area, or Ti83/84 normalcdf(low-zscore, hi-zscore)
The area to the left
of z = 0.75 is
0.2266.
Larson/Farber
Subtract to find the area of the
region between the two zscores:
0.8907  0.2266 = 0.6641.
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Practice
#1: Find the area under the standard
normal curve to the left of z = -0.99
Answer: 0.1611
0.99
z
0
#2: Find the area under the standard
normal curve to the right of z = 1.06
Answer: 0.1446
0
1.06
z
#3: Find the area under the standard
normal curve between z=-1.5 and 1.25
Answer: 0.8276
Larson/Farber
1.50
0
1.25
z
5.2 Probability and Normal Distributions
• If a random variable x is normally distributed, you can find the probability
that x will fall in a given interval by calculating the area under the normal
curve for that interval.
P(x < 600) = Area
μ = 500
σ = 100
x
μ =500 600
Example:
Normal Distribution
μ = 500 σ = 100
P(x < 600)
Standard Normal Distribution
μ=0 σ=1
x   600  500
z

1

100
P(z < 1)
x
μ =500 600
Larson/Farber
Same Area
P(x < 600) = P(z < 1)
z
μ=0 1
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Example1: Finding Probabilities for
Normal Distributions
A survey indicates that people use their computers an average of 2.4 years before
upgrading to a new machine. The standard deviation is 0.5 year. A computer owner is
selected at random. Find the probability that he or she will use it for fewer than 2 years
before upgrading. Assume that the variable x is normally distributed.
Normal Distribution
μ = 2.4 σ = 0.5
z
P(x < 2)
Standard Normal Distribution
μ=0 σ=1
2  2.4
P(z < -0.80)
x

 0.80

0.5
0.2119
z
x
2 2.4
-0.80 0
P(x < 2) = P(z < -0.80) = 0.2119
Find the probability using your Ti-84/83 Calculator___________________
Larson/Farber
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Example2: Finding Probabilities for
Normal Distributions
A survey indicates that for each trip to the supermarket, a shopper spends an average of 45
minutes with a standard deviation of 12 minutes in the store. The length of time spent in the
store is normally distributed and is represented by the variable x. A shopper enters the store.
Find the probability that the shopper will be in the store for between 24 and 54 minutes.
Normal Distribution
μ = 45 σ = 12
P(24 < x < 54)
Standard Normal Distribution
x -  24 - 45
μ=0 σ=1
z1 

 -1.75

12
x -  54 - 45
P(-1.75 < z < 0.75)
z2 

 0.75

12
0.7734
0.0401
x
24
45 54
z
-1.75
0 0.75
P(24 < x < 54) = P(-1.75 < z < 0.75) = 0.7734 – 0.0401 = 0.7333
Find the probability using your Ti-84/83 Calculator_________________
Example3: Finding Probabilities for
Normal Distributions
Find the probability that the shopper will be in the store more than 39 minutes.
(Recall μ = 45 minutes and σ = 12 minutes)
Normal Distribution
μ = 45 σ = 12
P(x > 39)
Standard Normal Distribution
μ=0 σ=1
z
x-

39 - 45

 -0.50
12
P(z > -0.50)
0.3085
z
x
39 45
-0.50 0
P(x > 39) = P(z > -0.50) = 1– 0.3085 = 0.6915
Find the probability using your Ti-84/83 Calculator__________________
If 200 shoppers enter the store, how many shoppers would you expect to be in
the store more than 39 minutes? 200(.6915) = 138.3 (approx. 138 shoppers)
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You Try this one!
Assume that cholesterol levels of men in the United States are normally
distributed, with a mean of 215 milligrams per deciliter and a standard
deviation of 25 milligrams per deciliter. You randomly select a man from the
United States. Draw pictures, use Z-scores and use your calculator to answer
the questions below:
1. A lower risk of heart attack is associated with a cholesterol level below 200.
What is the probability that the man’s cholesterol level is less than 200?
2. A moderate risk of heart attack is associated with a cholesterol level between
200 and 239. What is the probability that the man’s cholesterol level is
between 200 and 239?
3. A higher risk of heart attack is associated with cholesterol levels above 239.
What is the probability that the man’s cholesterol level is above 239?
(How would you do this using a complement?)
Answers: 1) .2743 2) .5572 3) .1685
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5.3 Finding a z-Score Given an Area
Example1: Find the z-score that corresponds to a cumulative area of 0.3632.
0.3632
z 0
z
TI 83/84
<2nd><DISTR>
3:InvNorm(.3632)
The z-score
is -0.35.
Larson/Farber 4th ed
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Finding a z-Score Given an Area
Example2: Find the z-score that has 10.75% of the distribution’s area to its
right.
1 – 0.1075
= 0.8925
0.1075
z
0
z
Because the area to the right is 0.1075, the cumulative area is 0.8925.
Locate .8925 in the Standard Normal Table. Or InvNorm (.8925)
Answer: Z-score = 1.24
Larson/Farber 4th ed
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Finding a z-Score Given a Percentile
Find the z-score that corresponds to P5 (or the 5th Percentile)
(Recall: 5th percentile refers to a value that is higher than 5% of the data)
The z-score for P5 is the same z-score that corresponds to an area of 0.05.
0.05
z
0
z
The areas closest to 0.05 in the table are 0.0495 (z = -1.65) and 0.0505 (z = 1.64). Because 0.05 is halfway between the two areas in the table, use the zscore that is halfway between -1.64 and -1.65. The z-score is -1.645.
OR
InvNorm (.05) = -1.645
Larson/Farber 4th ed
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Transforming a z-Score to an x-Score
To transform a standard z-score to a data value x in a given population, use the
formula :
x = μ + zσ
The speeds of vehicles along a stretch of highway are normally distributed,
with a mean of 67 miles per hour and a standard deviation of 4 miles per hour.
Find the speeds x corresponding to z-sores of 1.96, -2.33, and 0.
• z = 1.96:
x = 67 + 1.96(4) = 74.84 miles per hour (Above the mean)
• z = -2.33:
x = 67 + (-2.33)(4) = 57.68 miles per hour (Below the mean)
• z = 0:
x = 67 + 0(4) = 67 miles per hour (Equal to the mean)
Larson/Farber 4th ed
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Finding a Specific Data Value
Scores for a civil service exam are normally distributed, with a mean of 75 and
a standard deviation of 6.5. To be eligible for civil service employment, you
must score in the top 5%. What is the lowest score you can earn and still be
eligible for employment?
1 – 0.05
= 0.95
5%
0
?
z
x
invNorm (.95)
?
An exam score in the top 5% is any score above the 95th percentile. Find
the z-score that corresponds to a cumulative area of 0.95.
75
From the Standard Normal Table, the areas closest to 0.95 are 0.9495 (z = 1.64)
and 0.9505 (z = 1.65). Because 0.95 is halfway between the two areas in the table,
use the z-score that is halfway between 1.64 and 1.65. That is, z = 1.645
OR
Using the equation x = μ + zσ : x = 75 + 1.645(6.5) ≈ 85.69
The lowest score you can earn and still be eligible for employment is 86.