251solnM1 4/4/06 (Open this document in 'Page Layout' view!)
M. Continuous Distributions.
1. Introduction.
These are optional! 6.28-6.32 [6.29 – 6.33] (6.24 - 6.28)
2. Properties of the Normal Distribution.
Text 6.1-6.2, 6.4!, 6.5a-c [6.1, 6.2, 6.5a-e.] (6.1, 6.2, 6.5a-e.) M1 a-g, M2, M3.
3. Percentiles and Intervals about the Mean.
Text 6.5d, 6.6!, 6.8 [6.4, 6.5f-h, 6.8.] (6.4, 6.5f-h, 6.8.) M4, M1 h-j. Graded assignment 4. M8.
4. Normal Approximation to the Binomial Distribution.
6.41 [6.57*], M6, M7.
5. Normal Approximation to the Poisson Distribution.
? [6.60*], M5.
Exercise 6.60(Not in 8th edition): The number of cars arriving per minute at a toll booth is Poisson distributed with a mean of 2.5.
What is the probability that in any given minute: a. No cars arrive. b. Not more than 2 cars arrive? c. What is the approximate probability that in a ten minute period not more than 20 cars arrive? d.
What is the approximate probability that in a ten minute period between 20 and 30 cars arrive?
6. Review of Conditions for Approximation of One Distribution by Another.
This document contains solutions for Sections 2 - 3, except for M8, which is in 251solnM3.
--------------------------------------------------------------------------------------------------------------------------
This is a continuous distribution, so there is no distinction between ‘<’ and ‘  ’.
Exercise 6.1 in 9th: Make a diagram for each part!!!!
z ~ N (Normal, with a mean of 0 and a standard deviation of 1)
Exercise 6.1 in 10 th is a-d below: a) P
 z

1 .
57


P
 z

1 .
57 z

1 .
57


P
 z

0
 
0
 z

1 .
57


.
5

.
4418

.
9418
Your diagram should show a Normal curve with a center at zero indicated by a vertical line. 1.57
is above zero. Shade the entire area below 1.57. b) P
 z

1 .
84
  z

1 .
84


P
 z

0
 
0
 z

1 .
84


.
5

.
4671

.
0329
Your diagram should show a Normal curve with a center at zero. Shade the entire area above
1.84. It will be smaller than the area in the previous problem. c) P

1 .
57
 z

1 .
84


P

0
 z

1 .
84
 
0
 z

1 .
57


.
4671

.
4418

.
0253
Your diagram should show an area shaded between 1.57 and 1.84. Both are above zero. d) P
 z

1 .
57
  z

1 .
84


.
9418

.
0329
and 1.84 will not be shaded. Perhaps, a better way to do this is

1

P

1 .
57
 z

1 .
84


1

.
0253

.
9747 .

.
9747 .
On your diagram, only the area between 1.57
P
 z

1 .
57
  z

1 .
84

Exercise 6.2 in 10 th is e, f, h and i below: e) P


1 .
57
 z

1 .
84


P


1 .
57
 z

0
 
0
 z

1 .
84


.
4418

.
4671

.
9089 . Your diagram
will show a shaded area on both sides of zero. f) P
 z
 
1 .
57
  z

1 .
84


1

P


1 .
57
 z

1 .
84


1

.
9089

.
0911 .
g) The value of z with 50% of all probability above it. The point with 50% above it is called z
.
50
.
It is the median of this distribution. Since zero has .5 above it, z
.
50

0 .
h) The value of z with 2.5% of all probability above it. We now want the point
P
 z
 z
.
025


.
025 .
Make a diagram, showing z
.
025
, defined by z
.
025
with 2.5% above it and 97.5% below it.
Since 50% is below zero, z
.
025
must be above zero and have 97.5% - 50% = 47.5% between it
and zero. Mark the 50%, 47.5% and 2.5% as areas on your diagram. Go to the inside of your table
and find the number .4750. It will correspond to

.
4750
P
 z
 and
1 .
96

P
P
 z z


1 .
96
0



P
.
025
0
 z
.

1 .
96


.
5
 z
We conclude that
.
4750

1 .
96 .
z
.
025


.
0250
In other words
1 .
96
P

0
 z

. To check this find
1 .
96


251solnM1 4/4/06 (Open this document in 'Page Layout' view!) i) A symmetrical interval about zero containing 68.26% probability. Note that 68.26% divided
by 2 is 34.13%. One of these points must have 34.13% between it and zero and 50% - 34.13% =
15.87% above it. So we can say P

0
 z
 z
.
1587


.
3413 .
Make a diagram, showing an area between
 z
.
1587
and zero with a probability of .3413 and a second area between zero and z
.
1587 with a probability of .3413. If we find .3413 in the interior of the table, we find it corresponds to z
P


1 .
00 .

1 .
00
So
 z
P


0
0

 z


1 .
00
.
3413


.
3413 . But the symmetry of the distribution tells us that
. Thus the values z
.
1587

1 and
 z
.
1587
 
1 surround the requested interval.
To check this P


1 .
00
 z

1 .
00


P


1 .
00
 z

0
 
0
 z

1 .
00


.
3413

.
3413

.
6826 .
.
Exercise 6.2 in 9th: Make a diagram for each part!!!!
a) b) c) d) e) f)
P
P
P
P
P
P





 z z

0 z


1 .
34
1 .
17
 z

1 .
17

1 .
17

1 .
17



 
1 .
17 z
  z



 z
 z


1 .
34
1 .
17
.
3790
1 .
34 z






P
P
 z z


0
0
 

P
0
0

 z z

 z ~ N
1 .
34

1 .
17



.
5
.
5


.
4099
.
3790
This comes directly from the table.

1 .
17

0 .
50


P

 

1 .
17
P


1 .
17

 z z z


0
0

 
0

P

P

1 .
17
0
 z
0 .
50


 z

1 .
34
 z

0


.
5



.
0901
.
8790
.
3790
0



.
3790
.
3790

.
4099


.
1210
.
1915

.
7889

.
1875
Exercise 6.2 in 10th: Make a diagram for each part!!!!
a) b) c)
P
P
P


 z z


1 .
08

0 .
21

1 .
96

  z


P z
 z

0 .
21
1 .
08






0 .
21
P

 z
P
P


1 .
96
 z

0
 z
 
0

0
 


0 .
21 z

1 .
08

0 .
21

 z ~ N


P

 z


1 .
96
.
5
0



 z
.
3599
.
5


0

.
1401
.
0832
P

.
4168

0 .
21
 z
.
4750

.
0832

.
3918

0

 d) The value of z with 15.87% of all probability above it. We now want the point by P
 z
 z
.
1587


.
1587 .
Make a diagram, showing z
.
1587
, z
.
1587
, defined
with 15.87% above it and 84.13% below it. Since 50% is below zero, z
.
1587
, must be above zero and have 84.13% - 50% = 34.13% between it and zero. Mark the 50%, 34.13% and 15.87% as areas on your diagram. Go to the inside of your table and find the number .3413. It will correspond to

.
3413 and P
 z

1 .
00


.
5

.
3413 z


1 .
00 .
.
1587 .
In other words P
We conclude that

0
 z
 z
.
1587

1 .
00
1 .
00

.
To check this find
P
 z

1 .
00

P z

0
 
0
 z

1 .
00


.
5

.
3413

.
1587

251solnM1 4/4/06 (Open this document in 'Page Layout' view!)
Exercise 6.5a-e in 9 th : x ~ N

100 , 10

. Make diagrams. Do not make diagrams of x with zero in the
middle. Make up your mind! If you are diagramming x , put the mean in the middle; if you are diagramming z put zero in the middle. Remember z
 x



.
Exercise 6.5a in 10 th : a) P
 x

75


P

 z

75

100
10



P
 z
 
2 .
50


P


2 .
50
 z

0
  z

0


.
4938

.
5

.
9938
Make a diagram! If you make a diagram for x , it should be a Normal curve with 100 in the
middle. Shade the entire area above 75, which is below 100 all the way up to where the curve hits
the horizontal axis. If you make a diagram for z , it should be a Normal curve with 0 in the
middle. Shade the entire area above -2.5, which is below 0 all the way up to where the curve hits
the horizontal axis.
Exercise 6.5b in 10 th : b) P
 x

70


P

 z

70

100
10



P
 z
 
3 .
00


P
 z

0
 

3 .
00
 z

0


.
5

.
4987

.
0013
Make a diagram! If you make a diagram for x , it should be a Normal curve with 100 in the
middle. Shade the entire area below 70, which is below 100 all the way up to where the curve hits
the horizontal axis. If you make a diagram for z , it should be a Normal curve with 0 in the
middle. Shade the entire area above -3.0, which is below 0 down to where the curve hits the
horizontal axis. c) in 9 th edition – d in 8 th edition) P
 x

112


P

 z

112

10
100



P
 z

1 .
20
  z

0
 
0
 z

1 .
20


.
5

.
3849

.
1151
Make a diagram! If you make a diagram for x , it should be a Normal curve with 100 in the
middle. Shade the entire area above 112, which is above 100 all the way up to where the curve hits the horizontal axis. If you make a diagram for z , it should be a Normal curve with 0 in the
middle. Shade the entire area above 1.20, which is above 0 up to where the curve hits the
horizontal axis. d – c in 8 th edition)

P


2 .
50
 z

0
P

75
 x

85

 

1 .
50
 z


0

P



75

100
10
.
4938
 z

.
4332

85

100
10

.
0606



P


2 .
50
 z
 
1 .
50

Make a diagram! If you make a diagram for x , it should be a Normal curve with 100 in the
middle. Shade the area between 75 and 80, which are both below 100. If you make a diagram for z , it should be a Normal curve with 0 in the middle. Shade the area between -2.5 and -1.5, which
are both below 0.
Exercise 6.5c in 10 th : e ) P
 x

80

+

1

P


2 .
00

.
4772


.
3413
P
 x

110


1

P

80
 x

110

 z


1 .
00

.
8185
. Find
. So P

P x


2 .
00

80

+

P
 z x


1 .
00
110

1

P



80

100
10

2 .
00
 z

 z
0


110
P

10
0
100
 z



1 .
00


1

.
8185

.
1815
Make a diagram! If you make a diagram for x , it should be a Normal curve with 100 in the
middle. Shade the area below 80 and the area above 110. If you make a diagram for z , it should
be a Normal curve with 0 in the middle. Shade the area below -2 and the area above 1.00.

251solnM1 4/4/06 (Open this document in 'Page Layout' view!)
Problem M1: Assume that a.
b.
c.
d.
e.
f.
P
P
P
P
P
F





3
 x


6
 x

6
 x

0
 x

12
 x

  
12
12

P x
6

12

25



 x g.
F
12

6

P x


12

6

~ N ( 3 , 6 ) . Try the following: h. Find the 80th percentile. i. Find the 20th percentile. j. Find a symmetric region about the mean with 80% probability.
Solution: ( x ~ N ( 3 , 6 ) means
 
3 and
 
6 . We use the transformation z

.
Make Diagrams! a.
P

3
 x

12


P


3

3
6
 b.
P



P


6

1 .
50 x

 z
6



0
P



 
6
0
6


3 z z



12
6

3



P

0
 z

1 .
50


.
4332 z

0 .
50
6

6
3


 

0

P


1 .
50 z

1 .
50
 z

0 .
50
 
0
 z



.
4332

.
1915

.
6247
0 .
50
 x



.)

251solnM1 4/4/06 (Open this document in 'Page Layout' view!) c.
P

6


P

0
 z x


12

1 .
50

P


 
6
0

6

3 z

 z

12
0 .
50


6

3



.
4332
P

0 .
50

.
1915
 z

1 .
50

.
2417
 d.
P

0

P


 x
0 .
50

12
 z



0
P



0
P

6
0
3

 z z


12

1 .
50
6

3




P
.
1915


0 .
50

.
4332

 z

1 .
50
.
6247
 e.
P

P


12
0

 z x


25
3 .
67


P


12
6

 
0
 z
3

 z

1 .
50

25

6

3


.
4999

P

1 .
50

.
4332

 z

.
0667
3 .
67
 f.
F

P

 

P
 x

12
 z

0
 
0
 z


P

 z
1 .
50



12
6

.
5000
3




P
 z
.
4332


1 .
50

.
9332 g.
F

P

  z


1 .
50
P
 x
 
6


P

 z



P
 z

0
 
0


6
 z
6

3



1 .
50

P
 z

 
1 .
50
.
5000


.
4332

.
0668 h. Find the 80th percentile.
From the diagram, we want a point 30% above the median. For z this is z
.
20
so that P

0
 z
 z
.
20


.
3000 .
From the interior of the Normal table the closest we can come is P

0
 z

0 .
84


.
2995 .
So z
.
20

0 .
84 .
To get back to x , use the formula x
   z

, so that x
.
20
   z
.
20



3

P
 z
0 .
84

0 .
84



8 .
04
P
 z
. To check this

0
 
0
 z

P
 x
0 .
84


8 .

04


.
5000
P


 z
..

8
2995
.
04
6


3


.
7995

.
80 i. Find the 20th percentile.
From the diagram, we want a point 30% below the median. For z this is z
.80
so that P
 z
.
80
 z

0


.
3000 .
Since 0.84 is about 30% above
The median , -0.84 must be about 30% below the median. So
Therefore x
.
80
   z
.
80
 
3

0 .
84
 
 
2 .
04 z
.
80
. To check this
 
0 .
84
P

 x
P

 z

2 .
04


0

 
0
P


 z z



2 .
04
0 .
84
6



3


.
5000

P
 z


.
2995

0 .
84


.
2005

.
20
. j. Find a symmetric region about the mean with 80% probability.
From the diagram, we want two points
P
P
 z
.
10
 z
.
90
0
 z
 z
.
10
 z
 z
.
10




.
8000
.
4000 z
.
90 and
. The upper point,
, and by symmetry

.
. The interval for z
.
10
will have z
.
90
 x can then be written z
.
10
so that
 z
.
10
. From the interior of the Normal table the closest we can come is x
  
P

0 z
.
10

 z

3

1 .
28

1 .
28


 
.
3997

3

So
7 .
68 or -4.68 to 10.68. To check this

P


1 .
28
 z

1 .
28


2 P

0
 z
P


4 .
68

1 .
28


 x

10 .
68
2 (.
3997 )


.

P



7994
4 .
68
6

.
80

3
 z

10 .
68

6
3



251solnM1 4/4/06 (Open this document in 'Page Layout' view!)
Problem M2: : If x ~ N , find the probabilities for the following intervals: Below -3; -3 to 0; 0 to 3;
3 to 6; 6 to 9; above 9.
Solution: Make a diagram showing a Normal curve centered at 3 with vertical lines at -3, 0, 6 and 9. An easy way to do this problem is with cumulative probabilities, a method that can be used for any continuous distribution. This means that we will not have to look up each probability twice. We start by finding the areas below -3, 0, 3, 6 and 9 and then subtract them. Remember that , if we are working with standardized variables, F


0 .
30
  z
 
0 .
30

and that this type of probability can usually be found by looking up a value of z on the Normal table and either subtracting the probability that you find from 0.5 or adding 0.5.
As an example P


3
 x

0
  x

0
  x
 
3


P

 z

0

10
3 


P

 z


3

10
3 


P
 z
 
0 .
30 z
 
0 .
60


F


0 .
30
 

0 .
60

The process of subtracting each value from the value in front of it is called 'differencing the column.' The results are shown in the table below. Of course z
 x



 x

10
3 x z F
 
-3
0
3
6
9

-0.60 .5-.2257=.2743
-0.30 .5-.1179=.3821
0
0.30 .5+.1179=.6179
0.60 .5+.2257=.7257

.5000
1.0000


P
 x



3


3 x




P
P
P



0
3

 x x




P
P

6 x

 x
9




3
6
9



.
2743
0


.
3821



.
5000
.
6179
.
7257



1 .
0000


.
3821
.
5000
.
6179
.
7257
.
2743





1078
.
1179
.
1179
1078
.
2743
Problem M3: . I am a maker of videotapes. One of my clients has been getting returns of 6-hour tapes with complaints that they are less than six hours. The client proposes that each time a batch of tapes arrives, the client will take home three tapes and record on them. If any are less than six hours, the batch will be rejected. a. If the tapes are made with a mean length of 6.1 hours and a standard deviation of 0.1 hours and the normal distribution applies to the actual length of the tape, what is the chance that a given batch is rejected. b. To get the probability of rejection down to 1% what should the mean length be?
Solution: a) First find the probability that a tape picked at random is less than 6 hours. x ~ N

6 .
1 , 0 .
1

F
   x

6


P

 z

6 .
0

0 .
1
6 .
1 


P
 z
 
1 .
00


.
5000

.
3413

.
1587 . The test is to see if at least one of three tapes is less than 6 hours. We use the binomial distribution with q

1
 p

1

C
0
3

1

.
1587

.
8413 .
If y is the number of short tapes,

.
1587
 
.
8413

3 
1


.
8413

3 
1

.
5955

.
4045 .
n
P
 y

1



1

3 , p

.
1587 ,
P
 y

0


and
1

C
0
3 p
0 q
3

251solnM1 4/4/06 (Open this document in 'Page Layout' view!) b) This is fairly difficult. Don’t even think about doing it if you don’t understand all the previous problems.
We need a new mean length. Remember that P
 y

1


1

P
 y

0


1

C
0
3 p
0 q
3 
1
 q
3
, and we want 1
 q
3 
.
01 . This means that q
3 
.
99 , so that q
 3
.
99

.
9967 , and that p

1
 q

1

.
9967
 some unknown mean
.
0033 . If we look at part a), this means

. We want the value of
Normal curve with a mean of zero and z
0
so that z
0
below zero so that
F
   x

6


P

 z

6 .
0

0 .
1
 


.
0033 for
P
 z
 z
P
 z
0

 z

0
.
0033


. Make a diagram.
Show a
.
0033 and the probability between that P

0 z
0
and 0 is .5 - .0033 = .4967. If we look for .4967 on the inside of the Normal table, we will find
 z

2 .
72


.
4967 , so that z
0
 
2 .
72 this for the mean, multiply through by 0.1 to get
.
so we now know that
6 .
0
   
0 .
272 , or

6 .
0


0 .
1
6 .
0



0 .

2 .
72
272

. If we try to solve
6 .
272 .
To check this, note that if

.
5

.
4967 x ~ N

6 .
272 , 0 .
1

,

.
0033 .
This means that if number of short tapes,

1

.
9901

.
0099

3 ,
P
 y

1


1

P
 y

0


.
01 .
n
F p
  
 x

6

.
0033 and
 q
P

 z


1
 p
6 .
0


1
6 .
272
0 .
1



.
0033


P
 z

2 .
72
.
9967 , and
 y is the

1

C
0
3 p
0 q
3 
1

C
3
0

.
0033
 
.
9967

3 
1


.
9967

3
Exercise 6.4 in 9th: Make a diagram for each part!!!!
A lot of this is the same as 6.4 in 10 z ~ N
  a) b) c) d) e)
P
P

P

P

 z z


1 .
08

0 .
21

1 .
96

1 .
96
P


1 .
96
P

1 .
08



 z z z
  z




0

P
1 .
96 z
 z

0 .
21
 
1 .
08


1 .
08








0 .
21
P


1 .
96

0 .
21
.8349
.1151

P z
 z
 


0 z
0

 
 z



0
0
 


0 .
21
.
4750
 z

1 .
08


0 .
21


P

 z


1 .
96
.
0832


.
5
0


.
3918

 z
.
3599
.
5



.
1401
.
0832

0 .
21



.
4168 th .
By now, you should be able to get these answers without help. See me if you are still having trouble. f) The value of z with 50% of all probability below it. The point with 50% above it is called z
.
50
and is the same as the point with 50% below it. It is the median of this distribution. Since zero has
.5 above it, z
.
50

0 .
g) We want the .1587 fractile of the distribution. This point has 15.87% below it and 84.13%
above it and can be called z
.
8413
. Make a diagram showing 15.87% below z
.
8413
, 50% above zero
and 50% - 15.87% = 34.13% between z
.
8413
and zero. Of course
a probability of .3413 on the Normal table, it will be P

0
 z
 z
.
8413
is below zero. If you find
1 .
00

. So z
.
8413
  z
.
1587
 
1 .
00 .
h) The argument is pretty much the same. We want
P
 z
 z
.
1587


.
1587
Since the table says
.
P

0
 z

1 .
00


.
3413 , z
.
1587 z
.
1587
, which is defined by
A diagram will show that the probability between

1 .
00 . z
.
1587
and zero is .3413.

251solnM1 4/4/06 (Open this document in 'Page Layout' view!)
Exercise 6.5f-h in 9th: Make a diagram for each part!!!!
so that P
 x
 x
.
90


.
10 .
Make a diagram for x ~ N

100 f) We want the 10 th percentile of the distribution. This point is
, 10
 x
.
90
, defined by P
 x
 x
.
90


.
90 , z . Show a Normal curve with 0 in the middle and
a probability of .1000 below
should show that P
 z
.
90
 z
 z
.
90
which is a negative number and is equal to
0


.
4000 .
If we look for .4000 on probability part of the Normal
table, the closest we can come is .3997. The table says P

0
 z

1 .
28


.
3997
 z
.
10
.
The diagram
. So z
.
10

1 .
28
and z
.
90
 
1 .
28 . The formula to get from z to x is x
   z

, so x
.
90
   z
.
90


100

1 .
28

87 .
20 .
The last line of the t table gives z
.
10

1 .
282 , which would give a
slightly more accurate answer.
Exercise 6.5d in 10th: g) We want a symmetrical interval around the mean with a probability of 80%. Make a diagram for z . Show a Normal curve with 0 in the middle and an area between our two mystery points of
80% split into an area above zero of 40% and an area below zero of 40%. The diagram shows that
the probability above the higher number and the probability below the lower number are both
10%. We know from the previous section that 1.28 has about 10% above it and -1.28 has about
10% below it. We can summarize this by saying x
   z
.
10
 
100

1 .
28
 

100

12 .
8 or
87.2 to 112.8. h) We want x
.
70
, a point with 70% above it. This point must be below the mean, which has 50%
above it. Make a diagram for z . It will show 20% between
Because of symmetry
come to 20% is P

0 z
 z x
.
70
   z
.
70
 
100

.
70



0 .
52
 z
.
30
0 .
52


10


. A look at the Normal table shows us that the closest we can
.
1985
94 .
8 .
.
So z
30

0 .
52 and z
70
 
0 .
52 . This means that
If we check the last line of the t table, we can get a
slightly more accurate value z
.
30

0 .
524 .
z
.
70
and zero and 50% above zero.
Exercise 6.6 in 10 th edition: a. b.
P
P

 x x


43
42




.
9599
.
0228 x ~ N ( 50 , 4 ) c. The 5 th percentile.
Note that this value of z can be found using the t table. x
.
95

43 .
42 .
d. A symmetrical interval around the mean with a probability of 60%.
Note that this value of z can be found using the t table. The interval is 46.64 to 53.36.
Exercise 6.8 in 9th: Make a diagram for each part!!!!
x ~ N ( 50 , 12 )
Exercise 6.8a in 10th: a) P

34
 x

50


P


34

12
50
 z

50

50
12



P


1 .
33
 z

0


.
4082
Exercise 6.8b in 10th: b) P

P

34



1 .
33 x

 z
38



P


34
0
 


12
50
1 .
00

 z z


0

38


12
50
.
4082



P


1 .
33

.
3413

 z
.
0669
 
1 .
00
 c)1

P
1


30
 x
( P


1 .
67


60 z



1

P


0
 

30

12
50
0 .
83

 z

1

.
4525

.
2967

.
2508 z


60

50
0

)

12
1





1
.
4525


P


1 .
67
.
2967

 z

0 .
83


251solnM1 4/4/06 (Open this document in 'Page Layout' view!) d) P

30

P



1 .
67 x

 z
60


0

P


 
30


12
50
0 .
83

 z z


60

50
0


12
.
4525



P


1 .
67

.
2967

 z

.
7492
0 .
83

But this probability is also a proportion, so the number is
.
7492

1000


749 .
2 trucks.
Exercise 6.8c in 10th: e) It seems that the author wants
P
 x
 x
.
80


.
80 or P
 x
 x
.
80

 x
.
80
, the 20 th percentile of the distribution, defined by
.
20 since 80% of trucks will travel more than this mileage.
Make a diagram for z
80
  z
.
20 z showing a probability of .5000 above zero, a probability of .2000 below
and a probability of .3000 between
 z
.
20
and zero. Using the Normal table, the
closest we can come to P

0
 z
 z
.
20


.
3000 is P

0
 z

0 .
84


.
2995 , so
 z
.
20
 
0 .
84 and x
.
80
   z
.
80
 
50

0 .
84

39 .
92 thousand miles. f) The Instructor’s Solutions Manual provides the answers below.
The larger standard deviation makes the Z -values smaller.
(a) P (34 < X < 50) = P (– 1.60 < Z < 0) = 0.4452
(b)
(c)
(d)
(e)
P
P
(34 <
( X
X < 38) =
< 30) +
= 0.1815
P ( X
P (– 1.60 <
> 60) = P ( Z
Z < – 1.20) = 0.0603
< – 2.00) +
1000(1 – 0.1815) = 818.5 trucks
P ( Z > 1.00)
50 – 0.84(10) = 41.6 thousand miles or 41,600 miles
Problem M4: Assume x ~ N . Find: a. A 60% symmetrical interval about the mean. b. The 60th percentile. c. The 30th percentile. d. z
.23
e. x
.23
f. x
.52
Solution: See diagrams in handout. Make diagrams for all parts. a) A symmetrical interval about the mean with 60% probability.
The diagram: If you do a diagram for z , it will show two points, z
.
20
and z
.
80
. z
.
80
(which has 80% above it!) is below zero and z
.
20
is above. Since zero is halfway between these two points, the diagram will show 60% split between the two sides of zero, so that 30% is between z
.
80 and zero, and 30% is between zero and z
.
20
. The probability below z
.
80
and the probability above z
.
20
are both 20%.
From the diagram, we want two points upper point, z
.
20
will have P

0
 z
 z
.
20

 the Normal table the closest we can come is
.
3000
P

0 z
.
20
and
 z 0 .
84 z
.
80
so that P
 z
.
80
 z
 z
.
20


.
6000 . The
, and by symmetry



.
2995 z
.
80
  z
.
20
. From the interior of which implies that z
.
20

0 .
84 , and
our interval for x
   z
.
20
  z is -0.84 to 0.84. The interval for x can then be written
5

0 .
84
 

5

5 .
88 or -0.88 to 10.88.
Check:

2 P

0

P

 z

0 .
88
0 .
84


 x

10 .
88

2 (.
2995 )

P




.
5990
0 .
88
7

.
60

5
 z

10 .
88
7

5



P


0 .
84
 z

0 .
84


251solnM1 4/4/06 (Open this document in 'Page Layout' view!) b) Find the 60 th percentile.
The diagram: The diagram for z will show one point , z
.
40
(which has 40% above it!) which has 60% below it and is above zero, which is the 50 th percentile. Since zero has 50% below
it, the diagram will show only 10% between zero and
P
P


0
0
 z
From the diagram, we want one point
 z
 z

.
40
0 .
25


.
1000


.
0940 z
.
40
so that
0 .
25 z
.
40
.
P
 z
 z
.
40


.
6000 or
. From the interior of the Normal table the closest we can come is
, so the value of x can then be written x
   z
.
40
 
5

0 .
25
. This means that
 

5

1 .
75
 z
.
40
6 .
75 .

Check: P
 x

.
5

.
0940

6 .
75


.
5940

P

 z

.
60

6 .
75

7
5



P
 z

0 .
25


P
 z

0
 
0
 z

0 .
25
 c) Find the 30 th percentile.
The diagram: The diagram for z will show one point , z
.
70
  z
.
30
(which has 70% above it!) which has 30% below it and is below zero, which is the 50 th percentile. Since zero has 50% below
it, the diagram will show only 20% between zero and that P
 z
From the diagram, we want one point
.
70
 z

0


.
2000 . Because value of x can then be written x
   z
.
70
Normal table the closest we can come is P


0 z
.
70



 z
5 z z
.
30
.
70

,
 z
.
40
.
P


0
 z
.
30
so that
 z
 z
.
30


0 .
52
0 .
52 7

 


.
1985
5

3 .
64
P
 z
 z
.
70
.
2000
. So
 z
.
30
1 .
36

.


.
3000 . We can also see
. From the interior of the
0 .
52 or z
.
70
 
0 .
52 , and the
Check: To check this

P
 z

0
 

0 .
52
 z
P
 x


1 .
36


P

 z
0


.
5

.
1985

1 .
36

7
5 


.
3015

.

30
P
 z
 
0 .
52
 d) z
.23
The diagram: The diagram for z will show one point , z
.
23
,
(the 77 th percentile) which has 23% above it and is above zero, which is the 50 th percentile. Since zero has 50% below
it, the diagram will show 27% between zero and
From the diagram, we want one point z
.
23
. z
.
23
so that P
 z
From the interior of the Normal table the closest we can come is
 z
.
23
P

0

 z


.
2300
0 .
74

or

P

0
.
2704
 z
 z
.
23
. So z
.
23



.
2700
0 .
74 .
. e) x
.23
The diagram: The diagram for
In part d) we found that z is the same diagram that is used for part d) z
.
23

0 .
74 so the value of x can then be written

5

0 .
74

5

5 .
18

10 .
18 . x
   z
.
23

Check: P
 x

.
5

.
2704

10 .
18

.
2296



.
23
P

 z

10 .
18
7

5



P
 z

0 .
74


P
 z

0
 
0
 z

0 .
74


251solnM1 4/4/06 (Open this document in 'Page Layout' view!) f) x
.52
The diagram: The diagram for z will show one point , z
.
52
  z
.
48
(the 48 th percentile!) which has 52% above it (and 48% below it) and is below zero, which is the 50 th percentile. Since zero has 50% below it, the diagram will show only 2% between zero and that P
 z
From the diagram, we want one point
.
52
 z

0


.
0200 . Because value of x can then be written x
   z
.
52
Normal table the closest we can come is z
P


0
.
48



 z
5 z z
.
48

.
52
,

P


0
 z
0 .
05
0 .
05 7
 z
.
52
.
.
48
so that
 z
 z
.
52




.
0199
5

0 .
35
P
 z
 z
.
52
.
0200
. So
 z
.
48
4 .
65

.


.
5200 . We can also see
. From the interior of the
0 .
05 or z
.
52
 
0 .
05 , and the
Check: To check this

P


0 .
05
 z

P
 x

4 .
65


P

 z
0
  z

0


.
5

.
0199

4 .
65

7
5



.
5199


.
52
P
 z
 
0 .
05
