Transcript Density Curves and the Normal Distribution

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.
• Median: the equal-areas point, the point that
divides the area under the curve in half.
• Mean: the balance point, at which the curve
would balance if made of solid material.
Continuous Random Variable Ex 1.
A= .4(1) =0.4
0.4
0.5
What percent of the observations lie below 0.4?
40%
Continuous Random Variable Ex 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.
Normal Distributions
• Symmetric, single-peaked, and mound-shaped
distributions are called normal distributions.
• Normal curves: N(μ, σ)
– Mean = median
– The mean m and standard deviation s completely
determine the shape
Examples of Normal Curves:
The 68-95-99.7 Rule (aka Empirical Rule)
• A normal distribution with mean (μ) and standard
deviation (σ).
– 68% of the observations fall within (1σ) of (μ).
– 95% of the observations fall within (2σ) of (μ).
– 99.7% of the observations fall within (3σ) of (μ).
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%
Why are normal distributions important?
• Good descriptions of real data such as distributions
from tests taken by many people (ie. SAT exams and
Psychological tests), repeated careful measurements
of same quantity, characteristics of biological
populations.
• Good approximations of chance outcomes (ie. Coin
toss).
• Many statistical inference procedures based on
normal distributions work well for other roughly
symmetric distributions.
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 Lanier 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 Lanier 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 Lanier’s test was 70 pts.
• Mean score on Lanier’s test was 50.5 pts
while our test was 77.2 pts.
• Standard deviation on Lanier’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?
Lanier’s distribution
Iliana
50.5 55.8 61.1 66.4
Lanier’s and Hendrickson’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
z-scores:
• Tells us how many standard
deviations the original observation
falls away from the mean.
• Observations larger than the mean are
positive when standardized.
• Observations smaller than the mean
are negative when standardized.
Table A: front cover of textbook
• All normal distributions are the same when we
standardize them...
– Therefore we can use the same table to find the area
under the curve.
• The table entry for each value z represents the
area under the curve to the left of the z.
• You must subtract from 1 to find the area under
the curve to the right of the z.
http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_statistics.pdf
Ex 3. Find the proportion of the observations based on
the given…
• Z < 1.6
• Z ≤ -2.43
• Z > 1.31
• Z ≥ -.47
• -2.71 < Z ≤ 1.21
http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_statistics.pdf
Ex 4. Find the point z where…
• 65% of the observations fall below it.
• 83% of the observations fall below it.
• 32% of the observations fall above it.
http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_statistics.pdf
Ex 5. 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
 1667
.
z
3
Ex 6. 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.
a. 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
.
b. What percentile is the husband located in his department?
z  15
.
 93 percentile
rd
c. What percent of employees in the wife’s department earn
better than her?
z  125
.
P( z  125
. )  1.8944 .1056
d. 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.
e. If the husband wanted to earn in the 95th percentile, how
much should he earn per hour?
Need a z-score of 1.645!
14.60  12.80
z
 15
.
120
.
x  12.80
 1.645
1.20
x  12.80  1.974
x  $14.77
The husband will need to earn at least $14.77 to be in the 95th percentile.
Ex 7. The level of cholesterol in the blood is important because high cholesterol levels may
increase the risk for heart disease. The distribution of blood cholesterol levels in a large
population of people of the same age and sex is roughly normal. For 14 year old boys,
N(170, 30), levels above 240 mg/dl may require medical attention.
• What percent of 14 year old boys have more than 240 mg/dl of cholesterol?
• What percent of 14 year old boys have blood
cholesterol levels between 170 and 240
mg/dl?
Ex 8. The Beanstalk Club is limited to women and men who are very
tall. If women’s heights have a mean of 63.6 inches and a standard
deviation of 2.5 inches.
Find the cut off heights for the following chapters.
• The Travis County chapter wants to take women who are taller than
1.75 standard deviations above the mean.
• The Austin chapter wants to take only women in the top 8%.
• The Central Texas chapter wants to take women in the 88th
percentile.
http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_statistics.pdf
–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.