Transcript Describing Location in a Distribution

Describing Location in a
Distribution
2.2 Normal Distributions
YMS3e
2.2 Objectives
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Identify the main properties if the Normal curve
as a particular density curve.
List three reasons why Normal Distributions are
important in statistics.
Explain the 68-95-99.7 rule (the empirical rule).
Explain the notation N(μ,σ).
Define the standardized Normal distribution.
Use a table of values for the standard Normal
curve (Table A) to compute the proportion of
observations that are
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Less than a given z-score.
Greater than a given z-score.
Between two given z-scores.
2.2 Objectives
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Use a table of values for the standard Normal curve to
find the proportion of observations in any region given
any Normal distribution (i.e., given raw data rather than
z-scores)
Use a table of values for the standard Normal curve to
find a value with a given proportion of observations
above or below it (inverse Normal)
Identify at least two graphical techniques for assessing
Normality.
Explain what is meant by a Normal probability plot, use it
to help assess the Normality of a given data set.
Use technology to perform Normal distribution
calculations and to make Normal probability plots.
Properties of a Normal Distribution
x
• The mean, median, and mode are equal
• Bell shaped and is symmetric about the mean
• The total area that lies under the curve is one or 100%
Properties of a Normal Distribution
Inflection point
Inflection point
x
• As the curve extends farther and farther away from the
mean, it gets closer and closer to the x-axis but never
touches it.
• The points at which the curvature changes are called
inflection points. The graph curves downward between the
inflection points and curves upward past the inflection
points to the left and to the right.
Means and Standard Deviations
Curves with different means, same standard deviation
10 11
12 13 14
15 16 17 18 19
20
Curves with different means, different standard deviations
9 10 11 12 13 14 15 16 17 18 19 20 21 22
Empirical Rule
68%
About 68% of the area
lies within 1 standard
deviation of the mean
About 95% of the area
lies within 2 standard
deviations
About 99.7% of the area lies within
3 standard deviations of the mean
Determining Intervals
x
3.3 3.6 3.9 4.2
4.5 4.8 5.1
An instruction manual claims that the assembly time for a
product is normally distributed with a mean of 4.2 hours
and standard deviation 0.3 hour. Determine the
interval in which 95% of the assembly times fall.
95% of the data will fall within 2 standard deviations of the mean.
4.2 – 2 (0.3) = 3.6 and 4.2 + 2 (0.3) = 4.8.
95% of the assembly times will be between 3.6 and 4.8 hrs.
The Standard Normal Distribution
The standard normal distribution has a mean of 0 and a
standard deviation of 1.
Using z-scores any normal distribution can be
transformed into the standard normal distribution.
–4 –3 –2 –1
0 1
Larson/Farber Ch 5
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z
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Cumulative Areas
The
total
area
under
the curve
is one.
z
–3 –2 –1 0 1 2 3
• The cumulative area is close to 0 for z-scores close
to –3.49.
• The cumulative area for z = 0 is 0.5000.
• The cumulative area is close to 1 for z-scores close to
3.49.
Cumulative Areas
Find the cumulative area for a z-score of –1.25.
0.1056
–3 –2 –1 0 1 2 3
z
Read down the z column on the left to z = –1.25 and across to
the column under .05. The value in the cell is 0.1056, the
cumulative area.
The probability that z is at most –1.25 is 0.1056.
Finding Probabilities
To find the probability that z is less than a given value,
read the cumulative area in the table corresponding to
that z-score.
Find P(z < –1.45).
P (z < –1.45) = 0.0735
–3 –2 –1
0 1
2 3
z
Read down the z-column to –1.4 and across to .05. The
cumulative area is 0.0735.
Finding Probabilities
To find the probability that z is greater than a given
value, subtract the cumulative area in the table
from 1.
Find P(z > –1.24).
0.1075
0.8925
z
–3 –2 –1 0 1 2 3
The cumulative area (area to the left) is 0.1075. So the area
to the right is 1 – 0.1075 = 0.8925.
P(z > –1.24) = 0.8925
Larson/Farber Ch 5
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Finding Probabilities
To find the probability z is between two given values, find the
cumulative areas for each and subtract the smaller area from
the larger.
Find P(–1.25 < z < 1.17).
–3 –2 –1 0 1 2
1. P(z < 1.17) = 0.8790
3
z
2. P(z < –1.25) = 0.1056
3. P(–1.25 < z < 1.17) = 0.8790 – 0.1056 = 0.7734
Summary
To find the probability that z is less
than a given value, read the
corresponding cumulative area.
-3 -2 -1 0 1 2 3
z
To find the probability is greater
than a given value, subtract the
cumulative area in the table from 1.
-3 -2 -1 0 1 2 3
z
To find the probability z is between
two given values, find the cumulative
areas for each and subtract the smaller
area from the larger.
-3 -2 -1 0 1 2 3
z
Examples
 a.
Find the area under the curve to the left
of a z-score of –2.19.
 b.
Find the area under the curve to the left
of a z-score of 2.17.
Examples
c.
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Find the area under the standard normal
curve to the left of z = 2.13.
Examples
d.
Find the area under the standard normal
curve to the right of z = -2.16.
Probabilities and Normal Distributions
If a random variable, x is normally distributed, the
probability that x will fall within an interval is equal to the
area under the curve in the interval.
IQ scores are normally distributed with a mean of 100
and a standard deviation of 15. Find the probability that a
person selected at random will have an IQ score less
than 115.
100 115
To find the area in this interval, first find the standard
score equivalent to x = 115.
Z = 115 – 100 = 1
15
Probabilities and Normal Distributions
Normal Distribution
Standard Normal
Distribution
100 115
SAME
SAME
Find P(x < 115).
Find P(z < 1).
0 1
P(z < 1) = 0.8413,
so P(xCh <115)
= 0.8413
Larson/Farber
5
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Application
Monthly utility bills in a certain city are normally
distributed with a mean of $100 and a standard deviation
of $12. A utility bill is randomly selected. Find the
probability it is between $80 and $115.
Normal Distribution
P(80 < x < 115)
P(–1.67 < z < 1.25)
0.8944 – 0.0475 = 0.8469
The probability a utility bill is
between $80 and $115 is 0.8469.
From Areas to z-Scores
Find the z-score corresponding to a cumulative area of 0.9803.
z = 2.06 corresponds
roughly to the
98th percentile.
0.9803
–4 –3 –2 –1 0
1
2
3
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z
Locate 0.9803 in the area portion of the table. Read the
values at the beginning of the corresponding row and at
the top of the column. The z-score is 2.06.
Finding z-Scores from Areas
Find the z-score corresponding to the 90th percentile.
.90
0
z
The closest table area is .8997. The row heading is
1.2 and column heading is .08. This corresponds to
z = 1.28.
A z-score of 1.28 corresponds to the 90th percentile.
Finding z-Scores from Areas
Find the z-score with an area of .60 falling to its right.
.40
.60
z
0
z
With .60 to the right, cumulative area is
.40. The closest area is .4013. The row
heading is 0.2 and column heading is .05.
The z-score is 0.25.
A z-score of 0.25 has an area of .60 to its right. It
also corresponds to the 40th percentile
Finding z-Scores from Areas
Find the z-score such that 45% of the area under the
curve falls between –z and z.
.275
.275
.45
–z 0
z
The area remaining in the tails is .55. Half this area is
in each tail, so since .55/2 = .275 is the cumulative area
for the negative z value and .275 + .45 = .725 is the
cumulative area for the positive z. The closest table
area is .2743 and the z-score is 0.60. The positive z
score is 0.60.
From z-Scores to Raw Scores
To find the data value, x when given a standard score, z:
The test scores for a civil service exam are normally
distributed with a mean of 152 and a standard deviation of 7.
Find the test score for a person with a standard score of:
(a) 2.33
(b) –1.75
(c) 0
(a) x = 152 + (2.33)(7) = 168.31
(b) x = 152 + (–1.75)(7) = 139.75
(c) x = 152 + (0)(7) = 152
Larson/Farber Ch 5
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Finding Percentiles or Cut-off Values
Monthly utility bills in a certain city are normally distributed
with a mean of $100 and a standard deviation of $12. What is
the smallest utility bill that can be in the top 10% of the bills?
$115.36 is the smallest
value for the top 10%.
90%
10%
z
Find the cumulative area in the table that is closest to
0.9000 (the 90th percentile.) The area 0.8997 corresponds
to a z-score of 1.28.
To find the corresponding x-value, use
x = 100 + 1.28(12) = 115.36.