251y0011 10/11/00
Part I.
ECO251 QBA1
FIRST HOUR EXAM
OCTOBER 7, 2000
Name ____KEY___________
SECTION MWF 10 11 TR 11 12:30
Multiple Choice (10 points)
1.(D7-1) The major contribution of inferential statistics is that it
a. Allows us to take population information and make statements about samples.
b. Gives us a description of data contained in a sample.
c. Gives us a description of data contained in a population.
*d. Allows us to take sample information and make statements about the population.
e. None of the above.
2.(S-3) Debit balances owed in a retail store are an example of
a. Ordinal data.
b. Nominal data.
c. Interval data.
*d. Ratio data.
e. None of the above.
3. A used automobile dealer lists cars in the following classes. A - 100,000 miles or more on the odometer,
B - less than 100,000 miles on the odometer, C - Diesel. Are these three categories
a. Mutually exclusive?
*b. Collectively exhaustive?
c. Both mutually exclusive and collectively exhaustive?
d. Neither mutually exclusive or collectively exhaustive?
e. Can't tell with the information given.
4. (D7-9)If a distribution is skewed to the right, we can say that it is likely that
*a. Mean > median > mode
b. Median > mean > mode
c. Mode > median > mean
d. Mode > mean > median
e. Mode = mean = median (Most people got this backwards - make a diagram!)
5. A graph that connects points, each of which represents the frequency  f  is called a
a. Histogram
b. Ogive
*c. Frequency Polygon
d. Pie chart
e. None of the above
251y0011 10/11/00
Part II. Compute an appropriate answer, showing your work (except in a)) (15 Points maximum - if you do
more than 15 points, only your right answers will be counted.):
a) Fill in the following table (3)
Class
F
f rel
f
Solution:
50-59.99
60-69.99
70-79.99
80-89.99
90-99.99
Total
_
4
_
6
7
25
.12
__
__
__
_
__
Class
f
3
4
5
6
7
25
f rel
.12
.16
.20
.24
.28
1.00
50-59.99
60-69.99
70-79.99
80-89.99
90-99.99
Total
__
__
12
__
__
F
3
7
12
18
25
Note that n  25 .
b) Assume that we have sold 1000 life insurance policies in amounts between $5200 and $9800. If
this data is to be presented in eight classes, what intervals would you use? Explain your reasoning
using the appropriate formula and make a table showing the class intervals you would actually use.
(3)
9800  5200
 575 so use 600. This is only a suggestion. Any number somewhat above
8
575 will work.
Solution:
Class
A
B
C
D
E
F
G
H
From
5200
5800
6400
7000
7600
8200
8800
9400
To
5799.99
6399.99
6999.99
7599.99
8199.99
8799.99
9399.99
9999.99
c) (S-30)If a population of 1000 items with an unknown distribution has a mean of 12 and a
standard deviation of 1.2, what is the approximate minimum number of items that must be (i)
between 6 and 18? (ii) What is the maximum that can be above 18? (3)
x
6  12
18  12
 5 and
 5.
Solution: (i) If we use the formula k  z 
, we find that

1.2
1.2
According to the Chebyshef inequality, the minimum fraction of the data that must be between
1
1
  5 is 1  2  1   24 25 . 24 25 of 1000 is 960. (ii) The answer is the opposite to the
25
k
answer to (i). There are about 1000 - 960 = 40 items left over. All of these could be above 18.
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d) Do c) again assuming that the distribution is unimodal and symmetric.(2)
Solution: Since the Empirical Rule says that almost all points must be between   3 , we would
expect almost all of the 1000 points to be between 6 and 18 since these points are   5 , and we
would be quite surprised if even one point is above 18.
e) For the numbers 11.1, 13.2, 15.1 and 12.7, compute the i) Root-mean-square ii) Harmonic
mean, iii) Geometric mean (2 each)
x  52 .1 . This is not used in any of the following calculations and there is
Solution: Note that

no reason why you should have computed it!
(i) The Root-Mean-Square.
1
1
1
1
2
x rms

x 2  11 .12  13 .2 2  15 .12  12 .7 2  123 .21  174 .24  228 .01  161 .29   686 .75
n
4
4
4



1
n
 171 .6875 . So x rms 
x
2
 171 .6825  13 .103 .
(ii) The Harmonic Mean.
1
1
1 1 1
1
1
1  1

 



  0.090090  0.075758  0.066225  0.078740 
xh n
x 4  11 .1 13 .2 15 .1 12 .7  4

1
0.310813   0.077703 . So xh  1  1  12 .8947 .
1
1 0.077703
4
n
x
(iii) The Geometric Mean.


1
x g  x1  x 2  x3  x n  n  n
x 
4
11.113.21`5.112.7  4 28098 .1404  28098 .1404 
1
4
 28098 .1404 0.25  12.9470 .
Or
 
ln x g 
1
n
 ln( x)  4 ln 11.1  ln 13.2  ln 15.1  ln 12.7  4 2.40695  2.58022  2.71469  2.54160 
1
1
1
10.24346   2.56086 . So x g  e 2.56086  12 .9470 . I got the last result by putting 2.56086 into
4
the calculator and pressing 'inverse' and then 'ln x.'

Or
 
log x g 
1
n
 log( x)  4 log11.1  log13.2  log15.1  log12.7 
1
1
1.04532  1.12057  1.17898  1.10380   1 4.44868   1.11217 . So
4
4
x g  10 1.11217  12 .9470 . I got the last result by putting 1.11217 into the calculator and pressing

'inverse' and then 'log x.'
Notice that the original numbers and all the means are between 11.1 and 15.1. In spite of
everything that I said, there are many of you who think that: (i) You can find a sum of squares by
summing numbers and squaring the sum; (ii) You can find the sum of 1x by adding up the numbers
and taking the reciprocal; (iii) You can find an nth by dividing by n. I can only recommend a
remedial math class (unless, of course, you want to try listening in class and checking out the
homework very carefully.)
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Part III. Do the following problems (25 Points)
1. In a period of 7 days you make the following numbers of sales(in millions):
Day :
1
2
3
4
5
6
Sales: 9.2
10.2
9.2
11.2
19.5
12.2
Compute the following (assuming that the numbers are a sample):
a) Mean Sales (1)
b) The Median (1)
c) The Standard Deviation (3)
d) The 2nd Quintile (2)
7
13.2
Index x
 x  x 2
xx
x2
1
9.2
84.64
-2.9
8.41
2
9.2
84.64
–2.9
8.41
3 10.2 104.04
–1.9
3.61
4 11.2 125.44 or
-0.9
0.81
5 12.2 148.84
0.1
0.01
6 13.2 174.24
1.1
1.21
7 19.5 380.25
7.4
54.76
84.7 1102.09
0.0
77.22
Isn't it wonderful how predictable so many of you are! I strongly recommended that you compute
the variance by the computational formula in both this and the next problem. Many of you ignored me. Two
thirds of those who used the definitional formula got the problem wrong because they had not checked out
Solution: Compute the Following:
Note that x is in order
the method enough so that they knew what the formula meant.
you seem to have fooled yourselves into believing. Nor is
x
2
 x  x 
2
 x   84.7 as some of
equal to  x  x   84 .7  12 .1 .
is not
2
2
2
2
If you had tried these in any of the homework problems, you would have found that these tricks didn’t work.
Note that, to be reasonable, the mean, median and 2nd quintile must fall between 9.2 and 19.5.
n  6 ,  x  84 .7 ,
a) x 
 x  84.7  12.1
x
2
 1102 .09 ,
 x  x   0.00,  x  x 2  77.22 .
n
7
b) Just put the numbers in order and pick the middle number, 11.2.
Or formally: position  pn  1  a.b  .58  4.0
x1 p  xa  .b( xa1  xa ) so x1.5  x.5  x 4  .0( x5  x 4 )  11.2
c) s 2 
x
2
 nx 2
n 1

1102 .09  712 .12
 12 .87 or s 2 
6
 x  x 
n 1
2

77 .22
 12 .87
6
s  12.87  3.58748
d) The 2nd quintile has 40% below it. position  pn  1  a.b  .48  3.2
x1 p  xa  .b( xa1  xa ) so x1.4  x.6  x3  .2( x 4  x3 )  10.2  .2(11.2  10.2)  10.4
I warned you about quintiles - they are fifths, not fourths. This is an excellent warning! You can't answer a
question that you haven't read carefully!
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2. A bank finds that the amounts overdue on its credit cards are the following. . (Assume that the numbers
are a sample.) Are there reasons why so many of you (i) totally ignored the classes, (ii) decided that the
frequency column was both f and x , (iii) computed the
squaring it after I had specifically warned you not to?
amount (thousands)
a. Calculate the Cumulative Frequency (1)
b. Calculate The Mean (1)
c. Calculate the Median (2)
d. Calculate the Mode (1)
e. Calculate the Variance (3)
f. Calculate the Standard Deviation (2)
g. Calculate the Interquartile Range (3)
h. Calculate a Statistic showing Skewness and
Interpret it (3)
i. Make an histogram of the Data (Neatness
Counts!)(2)
frequency
0-$1.99999
$2.000-3.99999
$4.000-5.99999
$6.000-7.99999
$8.000-9.99999
$10.000 and up
fx 2 column by taking each value of fx and
80
40
30
30
20
0
Solution: x is the midpoint of the class. Our convention is to use the midpoint of 0 to 2, not 1.99999.
class
$0-$1.99999
$2.000-3.99999
$4.000-5.99999
$6.000-7.99999
$8.000-9.99999
n
 f  200 , 
 f x  x 
2
F
x
80 80
40 120
30 150
30 180
20 200
200
fx  740 ,
1.0
3.0
5.0
7.0
9.0
f
 1542.0, and

fx
fx3
fx 2
80
80
80
120 360 1080
150 750 3750
210 1470 10290
180 1620 14580
740 4280 29780
fx 2  4280 , and
fx 3

 f x  x 
3
x  x f x  x  f x  x 2 f x  x 3
-2.7 -216 583.2 -1574.64
-0.7
-28
19.6
-13.72
1.3
39
50.7
65.91
3.3
99 326.7 1078.11
5.3
106 561.8 2977.54
0 1542.0 2533.20
 29780 ,
f x  x   0,

 2533.20. Note that, to be reasonable, the mean, median and
quartiles must fall between 0 and 10. And no, I did not get the 1.0 in the x column by rounding 0.999995,
or, for that matter, by rounding anything else - Think!
a. Calculate the Cumulative Frequency (1): (See above) The cumulative frequency is the whole F column.
b. Calculate the Mean (1): x 
 fx  740  3.7
n
200
c. Calculate the Median (2): position  pn  1  .5201   100 .5 . This is above 80 and below 120, so the
 pN  F 
 .5200   80 
interval is 2-3.99999. x1 p  L p  
 w so x1.5  x.5  2  
 2  3.000
40
f


p


d. Calculate the Mode (1) The mode is the midpoint of the largest group. Since 80 is the largest frequency,
the modal group is 0 to 1.99999 and the mode is 1.000.
e. Calculate the Variance (3): s 2 
s2 
 f x  x 
n 1
2

 fx
2
 nx 2
n 1

4280  200 3.7 2
 7.74874 or
199
1542 .0
 7.74874
199
f. Calculate the Standard Deviation (2): s  7.74874  2.78366
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g. Calculate the Interquartile Range (3): First Quartile: position  pn  1  .25201  50.25 . This is above
 pN  F 
F  0 and below F  80 , so the group is 0 to 1.99999. x1 p  L p  
 w gives us
 f p 
 .25200   0 
Q1  x1.25  x.75  0  
 2  1.250 .
80


Third Quartile: position  pn  1  .75201   150 .75 . This is above 150 and below 180, so the group is
 .75 200   150 
6.000 to 7.99999. x1.75  x.25  6  
 2  6.000 . IQR  Q3  Q1  6.000 1.250  4.750 .
30


h. Calculate a Statistic showing Skewness and interpret it (3):
n
k 3
fx 3  3x
fx 2  2nx 3  200 29780  33.74280  2200 3.73
(n  1)( n  2)
199 198 





 0.00507588 2533 .2  12.8582 .
or k 3 
or g 1 
n
(n  1)( n  2)
k3
s
3

 f x  x 
12 .8582
2.78366 3
3

200
2533 .2  12.8582
199 198 
 0.596121
3mean  mode  33.7  1.0

 2.9098
std .deviation
2.78366
Because of the positive sign, the measures imply skewness to the right.
i. Make an histogram of the Data (Neatness Counts!)(2) A histogram is a bar graph of the frequency.
The first bar is between 0 and 2 on the x axis (or has a midpoint at 1) and has a height of 80.
or
Pearson's Measure of Skewness SK 
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