251solnG3a 1/31/08
G. Measures of Dispersion and Asymmetry.
1. Range
Downing & Clark, problem 7 above (Use data to find IQR). Review solutions and terms on page 41 (36 in 3 rd ed.) of Downing &
Clark.
2. The Variance and Standard Deviation of Ungrouped Data.
Text exercises 3.1b, 3.2b, 3.6, 3.37, 3.24 [3.1b, 3.2b, 3.7, 3.37, 3.23] (3.1b, 3.2b, 3.7, 3.23, 3.33)
3. The Variance and Standard Deviation of Grouped Data.
Text exercises 3.28, 3.30 (3.68, 3.70) (work 3.30 in thousands), Downing & Clark pg 42 or 37, problems 6,7 (Find sample standard
deviation – hint: run problem 6 in hundreds) (Note that you can use the Excel or Minitab techniques in the graded assignment to
compute and sum the
fx
and
fx 2
columns in problems 6 and 7. ), Problems G1, G2. Graded Assignment 1
4. Skewness and Kurtosis.
Find the standard deviation, coefficient of variation and measures of skewness in Text problem 3.1, 3.2. Problems G3A, G4 (See
251wrksht).
5. Review
a. Grouped Data.
b. Ungrouped Data.
This problem is in Section 4, but you were asked to do this problem on the computer.
Problem G3:
Class
0- 4.9
5- 9.9
10-14.9
15-19.9
20-24.9
25-29.9
30-34.9
35-39.9
40-44.9
45-49.9
50-54.9
55-59.9
Consider the following sample:
x
F
f
1
0
3
7
15
16
12
11
9
9
6
1
xf
90
Use computational formulas:
a. Complete the cumulative frequency under F .
b. Calculate the mean.
c. Calculate the median.
d. Calculate the mode.
e. Calculate the variance.
f. Calculate the interquartile range.
g. Calculate the standard deviation.
h. Calculate a statistic showing skewness.
i. Show all the data presented on a histogram with six class intervals.
j. Put a box plot below the histogram.
Now repeat Problem G3 using definitional formulas.
x2 f
x3 f
251solnG3a 1/31/08
Solution: Fill in the table. Note that the conventional way of writing the headings is Class, x , f , F, fx ,
fx2 and fx3 .
Class
0- 4.9
5- 9.9
10-14.9
15-19.9
20-24.9
25-29.9
30-34.9
35-39.9
40-44.9
45-49.9
50-54.9
55-59.9
x
f
F
(midpoint)
2.5
7.5
12.5
17.5
22.5
27.5
32.5
37.5
42.5
47.5
52.5
57.5
1
0
3
7
15
16
12
11
9
9
6
1
1
1
4
11
26
42
54
65
74
83
89
90
x2 f
xf
x3 f
2.5
6.25
0.0
0.00
37.5
468.75
122.5
2143.75
337.5
7593.75
440.0
12100.00
390.0
12675.00
412.5
15468.75
382.5
16256.25
427.5
20306.25
315.0
16537.50
57.5
3306.25
2925.0
106862.50
So if x = 12.5 and f = 3, fx  312.5  37.5 , fx2  37.512.5  468.75 and
15.675
0.000
5859.375
37515.625
170859.375
332750.000
411937.500
580078.125
690890.625
964546.875
868218.750
19109.250
4252781.250
fx3  468.7512.5  5859.375 .
First, we use computational formulas.
f  n  90,
To summarize our table,
 fx

3
 fx  2925 .0 ,  fx
2
 106862 .5 and
 4252781 .250 .
a. Complete the cumulative frequency under F: (See above). We add down the
b. Calculate the mean:
x
f column.
 fx  2925 .0  32.5
n
90
c. Calculate the median: To get a measure of position in grouped data
 pN  F 
first use position  pn  1 , then use x1 p  L p  
 w to find the value. Here
 f p 
p  .5 . So pn  1  .591  45.50 . This location is above 42 and below 54, so use 30
 .590   42 
to 34.9. Then x.5  30  
5  31 .25 .
12


d. Calculate the mode:
The group 25-29.9 has a frequency of 16, which is the largest frequency. So the mode is 27.5, its
midpoint.
e. Calculate the variance:
s2 
 fx
2
 nx 2
n 1

106862 .50  90 32 .52
 132 .584
89
251solnG3a 1/31/08
f. Calculate the interquartile range:
For the first quartile position  pn  1  .2591  22.75 . This location is above 11 and below
 pN  F 
26, so use 20 to 24.9. Then, using x1 p  L p  
 w we find
 f p 
 .25 90   11 
x.75  Q1  20  
5  23 .83
15


For the third quartile pn  1  .7591  68.25 . This location is above 65 and below 74, so
 .7590   65 
x.25  Q3  40  
5  41 .39 .
9


IQR  Q3  Q1  41.39  23.93  17.56
use 40 to 44.9. Then, we find
g. Calculate the standard deviation:
s  variance  132.584  11.515 . Note also
std .deviation 11 .515

 0.354 .
the coefficient of variation C 
mean
32 .5
h. Calculate a statistic showing skewness: There are three possibilities:
n
fx 3  3x
fx 2  2nx 3
1) k 3 
(n  1)( n  2)




90
4252781 .250  332 .5106862 .50   290 32 .53
(89 )(88)
 146 .514 .
k
146 .514
146 .514

 0.094 .
2) g1  33 
3
1526
.640
s
132 .584



3) Pearson’s Measure of Skewness SK 
i.

3mean  mode 332 .5  27 .5

 0.096 .
std .deviation
11 .51
Only one of these three is needed.
Show all the data presented on a histogram with six class intervals. Note: Because there were
only 6 intervals, the first bar on the graph was for 0 - 9.9 and had a height of 1 + 0 =1. The
next bar was for 10 - 19.9 and had a height of 3 + 7 = 10.
j. Put a box plot below the histogram.
(Include a hand-drawn solution to i and j.)
251solnG3a 1/31/08
Now we do the problem using definitional formulas. Note how much bigger the table has to be!
Class
x
0- 4.9
5- 9.9
10-14.9
15-19.9
20-24.9
25-29.9
30-34.9
35-39.9
40-44.9
45-49.9
50-54.9
55-59.9
(midpoint)
2.5
7.5
12.5
17.5
22.5
27.5
32.5
37.5
42.5
47.5
52.5
57.5
Note:
f
1
0
3
7
15
16
12
11
9
9
6
1
90
xf
F
x  x
x  x
2.5
0.0
37.5
122.3
337.5
440.0
390.0
412.5
383.5
427.5
315.0
57.5
2925.0
1
1
4
11
26
42
54
65
74
83
89
90
-30
-25
-20
-15
-10
-5
0
5
10
15
20
25
-30
0
-60
-105
-150
-80
0
55
90
135
120
25
0
f
 x  x 2 f
 x  x 3 f
900
0
1200
1575
1500
400
0
275
900
2025
2400
625
11800
27000
0
-2400
-23665
-15000
-2000
0
1375
9000
30375
48000
15625
12750
 f  n  90,  fx  2925 .0 ,  f x  x 2  11800 ,  f x  x 3  12750.
e. Calculate the variance:
 f x  x 
2
s2 
n 1

11800
 132 .584
89
h. Calculate a statistic showing skewness:
n
90
12750   146 .514
k 3
f x  x 3 
89 88 
(n  1)( n  2)
Other calculations are the same as on the previous page.
