Answers to homework problems for Chapters 3 and 5
1.
Consider the following dataset
X = {1,2,3,3,4,4,4,4,5}
Find the mean, median, mode, range, MAD, variance and standard deviation of this set of data. Assume these are sample data. Compute by hand and also use
Excel.
Mode = 4 (more 4’s than other value)
Median : there are N=9 observations. The median should be observation
median = X
N

1

X
10

X
5

4
2 2
Mean (assuming population data)
 
1
N i
N 

1
X i

1
9

1

2

3

3

4

4

4

4

5


30
9

3 .
3333
Range = Maximum value – minimum value = 5 – 1 = 4
MAD

1
N i
N 

1
X i
  
1
9


1


4
3 .
3333

3 .

3333

4

3 .
3333
2


3 .
3333

4

3 .
3333
3


3 .
3333

4

3 .
3333
3

3 .
3333



5

3 .
3333
MAD

1
9

2 .
3333
 
1 .
3333
 
0 .
3333
 
0 .
3333

0 .
6667

0 .
6667

0 .
6667

0 .
6667

1 .
6667
MAD

8 .
6667
9

0 .
9630

 2 
1
N
 
X i
  
2

1
9




1






3

4
3 .
3333
 
2


3 .
3333
3 .
3333

 
4
 
4
3 .
3333
 
3


3 .
3333
3 .
3333

 
4
 
5
3 .
3333

2


3 .
3333
3 .
3333

2

2






2 
1
9






2 .
3333

0 .
6667



1 .
3333
0 .
6667

 
0 .
3333
0 .
6667

 
0 .
3333

2
0 .
6667

1 .
6667

2




2 
1 .
5
  
2 
1 .
225
The problem using Excel

2.
Suppose that you know that
 
30 and
 
3 for a particular population. Assume that the population is normally distributed. a.
Describe the data using the empirical rule (this means to use the empirical rule for 1, 2, and 3 standard deviations). b.
How many of the observations have values < 27? c.
How many of the observations have values between 24 and 30? d.
How many of the observations have values between 24 and 33?
Approximately
68%
95%
Almost all



Of the observations will lie in an interval







3

,
2

,

,






 
2

3

30
 
 

30
30
3 , 30



6 , 30
9 , 30
3

 


9
6
27 , 33
 


24 , 36
21 , 39

 b. 2.5% + 13.5% = 16% have values less than 27 c. 13.5% + 34% = 47.5% d.13.5% + 34% + 34% =81.5%
3. Use Chebychev’s theorem with k=2 and k=3 to describe the data for the population in question 2.
1 Lie in

  k

,
  k


For k=2
At least

1 k
2

 
2

,
 
2

 
30

6 , 30

6
 
24 , 36


1
1 k
2
 
1
1
2
2

1

1
4

3
4

0 .
75
[24,36]
For k= 3

1
1 k
2
 
1
1
3
2

1

1
9

8
9

0 .
89

 
3

,
 
3

 
30

9 , 30

9
 
21 , 39

[21,39]

4. A machine is producing defectives at a 10% rate. A sample of 5 items is taken from the machines output. Use the binomial formula to find a.
The probability that the sample will contain no defective items. b.
The probability that the sample will contain exactly 1 defective item. c.
The probability that the sample will contain exactly 2 defective items. d.
The probability that the sample will contain 2 or fewer defective items. e.
The probability that the sample will contain 3 or move defective items.
Call finding a defective item a success, then p = 0.1. n = 5. a
P ( X
 x )

C x n
( p ) x
( q )
( n
 x )
. C x n
P ( X
 x !
n !
 n
 x

!

0 !
( 5
5 !

0 )!

5
5 !
!

1
 x )

1 (.
1 )
0
(.
9 )
5 
0 .
5905 b.
P ( X
 x )

C x n ( p ) x ( q ) ( n
 x )
C x n
P ( X
 x !
n !
 n
 x

!

1 !
( 5
5 !

1 )!

5 !
( 1 )( 4 !
)

1 )

5 (.
1 )
1
(.
9 )
4 
0 .
3281

5 c.
P ( X
 x )

C x n
( p ) x
( q )
( n
 x )
C x n
P ( X
 x !
n !
 n
 x

!

2 !
( 5
5 !

2 )!


2 )

10 (.
1 )
2
(.
9 )
3
( 1

5 !
2 )( 1

2

3 )

0 .
0729

10 d.
P ( X
P ( X


2 )

2 )

P ( X

0 .
5905
0 )


P ( X
0 .
3281


1 )

P ( X
0 .
0729


2 )
0 .
9915 e.
P ( X

3 )

1

P ( X

2 )

1

0 .
9915

0 .
0085

5.
Work problem 4 using Excel.
P(X=0)
P(X=1)
P(X=2)
P(X<=2)
P(X>=3)
0.5905 =BINOMDIST(0,5,0.1,FALSE)
0.3281 =BINOMDIST(1,5,0.1,FALSE)
0.0729 =BINOMDIST(2,5,0.1,FALSE)
0.9914 =BINOMDIST(2,5,0.1,TRUE)
0.0086 =1-BINOMDIST(2,5,0.1,TRUE)
6.
The reliability of an electrical fuse is the probability that a fuse, chosen at random from a production line, will function under the conditions for which it has been designed. A random sample of 1,000 fuses was tested and x = 27 defectives were observed. Calculate the probability of observing 27 or more defectives if the reliability is 0.98 (call S = finding a defective, then P(S) = 0.02.). Use Excel.
1-P(X<=26) 0.075814 =1-BINOMDIST(26,1000,0.02,TRUE)
7. Use the standard normal (Z) table to find the following probabilities. a. (

1.22)

?
b. P
 
1.23
Z 2.34
 
?
c.
  
1.34
 
?
d.
 
2.11
 
?
8.
Solve problem 1 using the NORMDIST() function in Excel. Hand in a copy of the Excel worksheet.
P(Z<1.22)
P(-1.23<Z<2.34)
P(Z> -1.34)
P(Z> 2.11)
0.8888 = NORMDIST(1.22, 0, 1, TRUE)
0.8810 = NORMDIST(2.34 ,0 ,1 , TRUE) - NORMDIST(-1.23, 0, 1, TRUE)
0.9099 = 1 - NORMDIST(-1.34, 0, 1, TRUE)
0.0174 = 1 - NORMDIST(2.11, 0, 1, TRUE)

9.
Use the standard normal table to find the indicated value of Z. a.
  z

 
0.10
b.
  z

 
0.80
c.
  z

 
0.20
d. P
  z

/ 2
Z z

/ 2
 
0.90
e. P
  z

/ 2
Z z

/ 2
 
0.95
Using the Norminv() functin in Excel
10.
IQ tests are designed to be normally distributed with mean 100 and standard deviation 15. Find the proportion of the population that has a. an IQ between 95 and 110.
Z
Z
P



95
X






15

0.67
X


X


110
15
 
 
0.33

P

0.33
0.67
   
0.3779
b.
an IQ less than 88.
Z

X





88
 
 
0.8
15
 
0.8)

0.2119

c.
an IQ greater than 135.
Z

X

 

135




2.33
15
(

2.33)
  
0.0099
11. Since the enactment of federal legislation on aerosol can propellants, manufacturers of hairsprays and deodorants have been introducing liquid sprays.
An new machine for filling cans of liquid hairspray can be set for any average fill.
If the amount of fill is normally distributed around a setting with a standard deviation of 0.05 oz, what setting will cause 90% of the cans to contain 12 or less ounces of liquid.
 
0.05
(
 x

)


 z



0.90
z


1.28
Z

X



or 1.28

12
 
0.05

=11.94

12. According to some sources, 50% of all loans extended by consumer finance associations are taken for the purpose of consolidating existing bills. Suppose that this is true. If so, find the probability that exactly 45 of 100 loans randomly selected from the files of a consumer loan agency were extended for the purpose of debt consolidation. Find the probability that 45 or more of the 100 loans were extended for the purpose of debt consolidation.
This is a binomial problem—the loan if for the purpose of consolidating existing bills or it is not. Call a Success a loan made for the purpose of consolidating existing bills. Then p=P(S) =0.5. Because of the size of n , it is unlikely that the binomial formula would be easy to use.
Check to see if the normal approximation to the binomial is possible np nq

100 0.5

50

100 0.5

50
So the approixmation should be acceptable.
  np

100 0.5

50
  npq

45



100 0.5 0.5

5
P

44.5

X

45.5

Z

X





5
 
0.9
Z

X
 
  
1.1
P
5
( 1.1
Z 0.9)

0.1841 0.1357

0.0484
Using Excel

And using Excel
  np

100 0.5

50
 
Z

Z
 npq

100 0.5 0.5

5

45
   
44.5

X




  
0.9
X



5
5
 
1.1
 
1.1) 1 0.1357

0.8643