Download 4630 Problem Set 1 answers

4630 Problem Set 1: SOLUTIONS
1.
a.
b.
There should be 6 classes.
Since the largest observation is 3.2 and the smallest is 1.5, the class
interval should be approximately (3.2 – 1.5)/6, or 0.28. For ease of
calculation, let us use 0.3 as the class interval.
c.
Class ranges
1.45-1.749
1.75-2.049
2.05-2.349
2.35-2.649
2.65-2.949
2.95-3.249
d.
midpoint
1.6
1.9
2.2
2.5
2.8
3.1
f
11
11
7
6
4
1
f/n
0.275
0.275
0.175
0.150
0.100
0.025
cum. f/n
0.275
0.550
0.725
0.875
0.975
1.000
(0.5)(40)  11
(0.3)  1.9955
11
(0.75)(40)  29
upper Q  2.35 
(0.3)  2.4
6
(0.25)(40)  0
lower Q  1.45 
(0.3)  1.7227
11
median  1.75 
IQR = 2.4 – 1.7227 = 0.6773
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e.
 1
1 6
X    f j X j   1.611  1.911  2.27   2.56  2.84  3.11
n  j 1
 40
83.2

 2.08
40
The actual mean (based on raw data) is 2.1.
2

1  6

2
S 
 fj Xj X 

n  1  j 1



1 
 111.6  2.08 2  111.9  2.08 2  72.2  2.08 2  62.5  2.08 2  42.8  2.08 2  13.1  2.08 2 

39 
1
 7.164  0.1837
39


So S = 0.4286. The actual variance and standard deviation (based on raw data)
are 0.2113 and 0.4597, respectively.
2.
The Citizens Banking Company is studying the number of times a particular ATM
is used each day. The following table lists the number of times the machine was
used over each of the last 30 days:
83
63
95
a.
64
80
36
76
73
61
84
68
59
54
52
84
75
65
95
59
90
47
70
52
87
61
77
60
What is the mean number of times the machine was used each day?
X
b.
84
84
78
1
1
 X i  (2116 )  70.5333
30
30
What is the modal number of times the machine was used each day?
The most commonly occurring value is 84.
c.
What is variance and standard deviation?
39
S2 
1
2 1
 ( X i  X )  6373 .467  219.7747
29
29
S = 14.8248
3.
a.
b.
c.
Tabulate and graph the relative frequency distribution.
On the average, how many calls are there per minute?
Calculate the variance and standard deviation.
X
1
2
3
4
5
f
19
20
7
2
2
f/n
0.38
0.40
0.14
0.04
0.04
25
20
15
Series2
10
5
0
1
2
3
4
5
1
(1)(19)  (2)(20)  (3)(7)  (4)(2)  (5)(2)  98  1.96
50
50
1
 f X  X 2  f X  X 2  ... f X  X 2 
S2 
 1 1

2 2
n n
n 1 

1
1

19(1  1.96) 2  20(2  1.96) 2  ...  2(5  1.96) 2 
51.92  1.0596
49
49
X


S = 1.0294
40
4.
The following data are the monthly rental prices for a sample of 10 unfurnished 1bedroom apartments in Denton and a sample of 10 unfurnished apartments in
Lewisville:
Denton Lewisville
$680
$460
$369
$499
$539
$489
$559
$664
$644
$479
$550
$425
$699
$835
$700
$670
$610
$499
$608
$655
SOURCE: Apartments.com, 8/10/05
a.
For each set of data compute the mean, median, range, interquartile range,
and standard deviation. NOTE: do not group these data.
statistic
mean
median
upper quartile
lower quartile
range
st. dev
b.
Denton
$538.20
$519.00
$649.00
$474.25
$311.00
$99.87
Lewisville
$625.10
$632.50
$699.25
$537.25
$410.00
$115.54
What can be said about unfurnished 1-bedroom apartments renting in
Denton versus Lewisville?
The sample average rent is higher in Lewisville, but the variation is
also greater. This makes it not totally clear that the true population
average rent is higher in Lewisville.
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