Transcript File

The Normal Distribution
Chapter 2
Continuous Random Variable
• A continuous random variable:
– Represented by a function/graph.
– Area under the curve represents the proportions of
the observations
– Total area is exactly 1.
• How do we locate the median for a continuous
random variable? the mean?
• The median is the value that divides the graph
into equal area while the mean is the “balance”
point.
Continuous Random Variable 1
A= .4(1) =0.4
0.4
0.5
What percent of the observations lie below 0.4?
40%
Continuous Random Variable 2
A= 1.4(.5) =0.7
0.6
What proportion of the observations lie above 0.6?
Notice, to find proportion for observation above,
we can use the complement rule.
Continuous Random Variable 3
• Where is the mean and median?
• How will the curve change as s changes?
Normal Distributions
• Symmetric, single-peaked, and mound-shaped
distributions are called normal distributions
• Normal curves:
– Mean = median
– The mean m and standard deviation s completely
determine the shape
• Fathom
The Normal Curve
• Will finding proportions work different than
previous random variable examples?
• Empirical Rule Discovery
68% of observations fall within 1s of m
95% of observations fall within 2s of m
99.7% of observations fall within 3s of m
68-95-99.7 Rule
Applet
68-95-99.7 Rule
34%
.15%
2.35%
Applet
13.5%
34%
13.5%
2.35%
Percentiles?
16th
34%
2.35%
13.5%
50th
84th
34%
13.5%
2.35%
What’s Normal in Statistics?
• Normal distributions are good descriptions for real
data allowing measures of relative position to be
easily calculated (i.e. percentiles)
• Much of statistical inference (in this course)
procedures area based on normal distributions
• FYI: many distributions aren’t normal
Distribution of dates is approximately
normal with mean 1243 and standard
deviation of 36 years.
1135
1171
1207
1243
1279
1315
1351
Assume the heights of college women are
normally distributed with a mean of 65
inches and standard deviation of 2.5 inches.
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are taller than 65 in.?
50%
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are shorter than 65 in.?
50%
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are between 62.5 in.
and 67.5 in.?
68%
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are between 60 in. and
70 in.?
95%
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are between 60 and
67.5 in?
68%
13.5%
81.5%
57.5
60
62.5
65
67.5
70
72.5
What percentage of women are shorter than 70 in.?
50%
34%
13.5%
97.5%
57.5
60
62.5
65
67.5
70
72.5
Iliana’s Grade
• After 5 weeks of class Iliana must transfer
from a stat class at Pflugerville to this class.
Last week was the chapter 1 test in both
classes. Iliana scored a 61 out of 70. Let’s
say our test was out of 100 points. What
score should she be given?
Iliana’s claim
• Iliana claims that her test at Pflugerville was
harder than our test.
• Does your previous method of assigning a
grade take in consideration difficulty?
• If we have all of the data, what important
facts can we utilize to improve our
assignment of Iliana’s grade?
Important Facts
• Maximum possible on our test was 100 pts
while Pflugerville’s test was 70 pts.
• Mean score on Pflugerville’s test was 50.5
pts while our test was 77.2 pts.
• Standard deviation on Pflugerville’s test
was 5.3 pts while ours was 8.1 pts.
• Test scores from both high schools tend to
be normally distributed.
• How will we fairly assign Iliana’s score?
Pflugerville’s distribution
Iliana
50.5 55.8 61.1 66.4
Pflugerville’s and McCallum’s
distributions
Iliana’s score – class average
standard deviation
Iliana
Iliana
50.5 55.8 61.1 66.4
77.2 85.3 93.4 101.5
•How can we find Iliana’s relative position?
Formula
• What is the formula to find the relative
position for any distribution?
Iliana’s score – class average
standard deviation
z–score=
xm
s
1. Suppose as student has taken two quizzes in a statistics course. On
the first quiz the mean score was 32, the standard deviation was 8, and
the student received a 44. The student obtained a 28 on the second
quiz, for which the mean was 23 and the standard deviation was 3. If
test scores are approximately normal, on which quiz did the student
perform better relative to the rest of the class?
z
xm
s
First quiz:
Second quiz:
44  32
z
 15
.
8
28  23
 1.667
z
3
3. A married couple is employed by the same company. The husband
works in a department for which the mean hourly rate is $12.80 and the
standard deviation is $1.20. His wife is employed in a department where
the mean rate is $13.50 and the standard deviation is $1.80.
Relative to their departments, which is better paid if the husband earns
$14.60 and the wife earns $15.75?
Husband:
Wife:
14.60  12.80
z
 15
.
120
.
15.75  1350
.
z
 125
.
180
.
What percentile is the husband located in his department?
z  15
.
 93
nd
percentile
What percent of employees in the wife’s department earn
better than her?
z  125
.
P( z  125
. )  1.8944 .1056
What would the wife need to earn to match her husband’s
relative position?
Husband:
Wife:
14.60  12.80
z
 15
.
120
.
15.75  1350
.
z
 125
.
180
.
x  1350
.
 15
.
180
.
x  1350
.  2.7
x  $16.20
The wife would need to earn $16.20 to match the husband’s relative
position.
If the husband wanted to earn in the 95th percentile, how
much should he earn per hour?
Need a z-score of 1.65!
14.60  12.80
z
 15
.
120
.
x  12.80
 165
.
1.20
x  12.80  198
.
x  $14.78
The husband will need to earn at least $14.78 to be in the 95th percentile.
–2
P  x  14.24   .0228
2
P  x  59.60   .9772
P 14.24  x  59.60   .9772  .0228  .9544
5.5%
94.5%
89%
44.5%
z-score = –1.60
x  36.92
1.60 
11.32
z-score = 1.60
x  36.92
1.60 
11.32
The middle 89% of the data ranges from 18.81 to 55.03 ppb.
Is the Data Normal?
Check for 68-95-99.7 Rule
Check for the 68-95-99.7 Rule
Normal Probability Plot
Page 129: 43, 45, 47, 49, 53, 63, 65
Practice Problems
• The average SAT math score for the 50 states and Puerto Rico in 1992
was 497.4 with a standard deviation of 34.57. What is the probability
that a randomly selected states SAT math score is less than 540 points?
• The average number of frost days during April in Greenwich, England
is 2.92 days with a standard deviation of 2.69 days. What is the
probability of a randomly selected April having one or fewer days of
frost?