Transcript Document

Chapter Six
Normal Curves and
Sampling Probability
Distributions
Chapter 6
Section 3
Areas Under
Any Normal
Curve
Finding Z Scores When
Probabilities (Areas) Are Given
Find the indicated z score: if P(0 < z < zcutoff) = 0.3907
0.3907
P(0 < z < 1.23)
.3907
0
z=?
z = 1.23
Find the indicated z score:
if P(zcutoff < z < 0) = 0.1331
0.1331
P(-0.34 < z < 0)
0.1331
?=z 0
z = –0.34
Find the indicated z score:
if P(z < zcutoff) = 0.8554
0.8554
0.5000+0.3554
P(z < 1.06)
0.8554-0.5000
.8554
0.3554
0
z=?
z = 1.06
Find the indicated z score:
if P(z < zcutoff) = 0.0681
0.0681
0.5000 - 0.4319
P(z < -1.49)
0.0681
0.5000-0.0681
0.4319
z
z = -1.49
0
Find the indicated z score:
if P(z > zcutoff) = 0.01
0.01
0.5000 - 0.4900
P(z > 2.33)
0.5000- .0100
.01
0.4900
z
0
z = 2.33
Find the indicated z score:
if P(z < zcutoff) = 0.005
0.0050
0.5000 - 0.4950
P(z < – 2.575)
0.5000-0.0050
.005
.4950
z
z = – 2.575
0
Find the indicated z score:
if 1% is in the tail regions
Area A + Area B = .01
and
Area A = Area B
0.01
2(0.005)
2(0.5000 - 0.4950)
P (z < -2.575) or P(z > 2.575)
So, Area A = 0.005
and Area B = 0.005
1.0000 - 0.0100
0.9900
0.4950 + 0.4950
A
.4950
.4950
–z
z = -2.575
0
B
z
z = 2.575
Find the indicated z score:
if 5% is in the tail regions
Area A + Area B = .05
and
Area A = Area B
0.05
2(0.025)
2(0.5000 - 0.4750)
P (z < -1.96) or P(z > 1.96)
So, Area A = 0.025
and Area B = 0.025
1.0000 - 0.0500
0.9500
A
0.4750 + 0.4750
.4750
.4750
–z
z = -1.96
0
B
z
z = 1.96
Application of Determining
z Scores
The Verbal SAT test has a mean score of
500 and a standard deviation of 100. The
scores are normally distributed. A major
university determines that it will accept
only students whose Verbal SAT scores are
in the top 4%. What is the minimum score
that a student must earn to be accepted?
Application of Determining
z Scores
m = 500 and s = 100
0.0400
The cut-off score
is 1.75 standard
deviations above
the mean.
0.5000 - 0.4600
P ( z > 1.75 )
æ x - 500
ö
Pç
> 1.75 ÷
è 100
ø
P ( x - 500 > 175 )
P ( x > 675 )
0.0400
0.4600
0
z
z = 1.75
Students would need to score 675 or above.
Application of Determining
z Scores
The length of time employees have been working at a
specific company is normally distributed with a mean of
15 years and a standard deviation of 5.2 years. The CEO
has decided due the recent hard economic times that the
work force must be reduced by 5%. He decided to offer a
retirement incentive for the longest working employees
and to lay off a portion of the most recently hired
employees. If he is able to split the percentage evenly
between the two groups, what is his target range for the
years of employment?
Application of Determining
z Scores
m = 15 years and s = 5.2 years
0.0500
The cut-off score
is ±1.96 standard
deviations.
2(0.0250)
2(0.5000-0.4750)
0.4750
-z
0.0250
z = -1.96
0.4750
0
z 0.0250
z = 1.96
Application of Determining
z Scores
m = 15 years, s = 5.2 years, and z = ±1.96
0.0250
0.0250
0.5000 - 0.4750
0.5000 - 0.4750
P ( z < -1.96 )
æ x - 15
ö
Pç
< -1.96 ÷
è 5.2
ø
P ( x - 15 < -10.192 )
P ( x < 4.808 )
P ( z > 1.96 )
æ x - 15
ö
Pç
> 1.96 ÷
è 5.2
ø
P ( x - 15 > 10.192 )
P ( x > 25.192 )
The CEO should offer the retirement package to those employees
that have worked at the company for over 25.1920 years and he
must layoff employees that have worked for the company less than
4.8080 years.
Check for Normality
1. Make a histogram and it should be roughly "bell-shaped".
2. If the distribution is nomal there should be no more than 1 outlier.
Outlier < Q1 + 1.5IQR or Outlier > Q3 + 1.5IQR
3. The data should be symmetric. A Pearson's index greater than
1 or less than - 1 indicates a skewed set of data and the data is
not considered normal.
3( x - median )
Pearson's index =
s
4. Graph a Normal Quantile Plot. The data is considered to be normally
distributed if the data points roughly form a straight line.
Normal Quantile Plot
Place your data points into L1
Press STATPLOT
Select option 1
Select ON
Type: Select the 6th graph
Data List: L1
Data Axis: Select Y
Mark: (the box)
Introduction
Packet
Questions
1. If x is a normally distributed variable with a
mean of 30 and a standard deviation of 6, find
the following probabilities:
P ( x ³ 30 )
30 - 30 ö
æ
Pç z ³
÷
è
6 ø
0ö
æ
Pç z ³ ÷
è
6ø
\
0.00
P ( z ³ 0)
0.5000
1. If x is a normally distributed variable with a
mean of 30 and a standard deviation of 6, find
the following probabilities:
P ( x ³ 36 )
36 - 30 ö
æ
Pç z ³
÷
è
6 ø
æ
Pç z ³
è
6ö
÷
6ø
P ( z ³ 1)
1.00
0.5000 - 0.3413
0.1587
1. If x is a normally distributed variable with a
mean of 30 and a standard deviation of 6, find
the following probabilities:
P ( x ³ 18 )
18 - 30 ö
æ
Pç z ³
÷ø
è
6
-12 ö
æ
Pç z ³
÷ø
è
6
-2.00
P ( z ³ -2 )
0.5000 + 0.4772
0.9772
1. If x is a normally distributed variable with a
mean of 30 and a standard deviation of 6, find
the following probabilities:
P ( 24 £ x £ 39 )
39 - 30 ö
æ 24 - 30
Pç
£z£
÷
è 6
6 ø
9ö
æ -6
Pç
£z£ ÷
è 6
6ø
P ( -1 £ z £ 1.5 )
-1.00
1.50
0.3413 + 0.4332
0.7745
2. Determine the z scores that produce the
following probabilities.
a. 0.20 lies to the right of the z-score
0.2000
0.5000 - 0.3000
P ( z ³ 0.84 )
0.84
2. Determine the z scores that produce the
following probabilities.
b. 0.40 lies to the right of the z-score
0.4000
0.5000 - 0.1000
P ( z ³ 0.25 )
0.25
2. Determine the z scores that produce the
following probabilities.
c. 0.875 lies to the right of the z score.
0.8750
0.5000 + 0.3750
P ( z ³ -1.15 )
-1.15
2. Determine the z scores that produce the
following probabilities.
d. 0.6328 lies between z and -z.
0.6328
0.3164 + 0.3164
P ( -0.90 £ z £ 0.90 )
-0.90
0.90
3. Consider the following data set: to answer
the following questions.
a. Make a histogram for this data
{1, 5, 2, 3, 4, 3, 3, 4, 4, 3, 2 }
3. Consider the following data set: to answer
the following questions.
b. Make a Box-n-Whisker plot for the data.
{1, 5, 2, 3, 4, 3, 3, 4, 4, 3, 2 }
<----+----+----+----+----+----+----+----+----+---->
0
1
2
3
4
5
6
7
8
3. Consider the following data set: to answer
the following questions.
c.
set
What is the Pearson’s Index for the
of
5, 2, 3, 4, 3, 3, 4, 4, 3, 2 }
{1,data?
x = 3.0909
PI
s = 1.1362
PI
median = 3
PI
PI
3( 3.0909 - 3)
=
1.1362
3( 0.0909 )
=
1.1362
0.2727
=
1.1362
= 0.2400
3. Consider the following data set: to answer
the following questions.
d.
Make a normal quantile plot for the
data.
3. Consider the following data set: to answer
the following questions.
e.
Interpret the results from parts a
through d.
The data appears to be normally distributed since:
1. The histogram seems to roughly fit the bell-shaped curve.
2. The Box-n-Whiskers plot does not have any outliers and
appears symmetrical.
3. The Pearson’s Index of Skewness is in the normal range of
-1 < PI < 1 since PI = 0.2400.
4. The Normal Quantile Plot (Normal Probability Plot)
produces a graph where the data points lie in a relatively
close line.
Homework Assignments
Pages 286 - 291
Exercises: 1, 5, 9, 13, 17, 21, 25, 29, 33, and 37
Exercises: 3, 7, 11, 15, 19, 23, 27, 31, 35, and 39
Exercises: 2, 6, 10, 14, 18, 22, 26, 30, 34, and 38
Exercises: 4, 8, 12, 16, 20, 24, 28, 32, 36, and 40