3.5 The Weighted Mean and Grouped Data:
Weighted Mean:
n
xw 
w x
i 1
n
i
i
w
i 1
.
i
Note: when data values vary in importance, the analyst must choose
the weight that best reflects the importance of each data value in the
determination of the mean.
Example 5:
The following are 5 purchases of a raw material over the past 3 months.
Purchase
Cost per Pound ($)
Number of Pounds
1
2
3
3.00
3.40
2.80
1200
500
2750
4
5
2.90
3.25
1000
800
Find the mean cost per pound.
[solutions:]
w1  1200, w2  500, w3  2750, w4  1000, w5  800.
and
x1  3.00, x2  3.40, x3  2.80, x4  2.90, x5  3.25.
Then,
5
xw 
w x
i 1
5
i
i
w
i 1
i
1200  3.00   500  3.40  2750  2.80  1000  2.90  800  3.25
1200  500  2750  1000  800
 2.96

1
Population Mean for Grouped Data:
m
g 
F
k 1
m
m
k
Mk
 Fk

F
k 1
k
Mk
,
N
k 1
where
Mk :
Fk :
the midpoint for class k,
the frequency for class k in the population,
m
N   Fk :
k 1
the population size.
Sample Mean for Grouped Data:
m
xg 

m
fk M k
k 1
m

k 1


k 1
fk
fk M k
n
,
where
fk :
the frequency for class k in the sample,
m
n   fk :
k 1
the sample size.
Population Variance for Grouped Data:
 Fk M k   g 
m
 g2 
2
k 1
N
2
Sample Variance for Grouped Data:
 f M
m
s g2 
k
k 1
 xg 
m
2
k
n 1

f
k 1
k
M k2  nxg2
n 1
Example 3 (continue):
(h) Based on the frequency table obtained in (g), compute the mean and variance for
the grouped data. Compare with the results in (a) and (b). (10%)
[solutions:]
xg 
5.5  3  15.5  4  25.5  4  35.5  4
 21.5
15
and
s
2
g
2
2
2
2

5.5  21.5  3  15.5  21.5  4  25.5  21.5  4  35.5  21.5  4

14
 125.71
Thus,
s g  11.212 .
The group mean and the group standard deviation are close to the original mean and
standard deviation.
Example 6:
The following are the frequency distribution of the time in days required to complete
year-end audits:
Audit Time (days)
Frequency
10-14
4
15-19
20-24
25-29
30-34
8
5
2
1
What is the mean and the variance of the audit time?
[solutions:]
3
f1  4, f 2  8, f 3  5, f 4  2, f 5  1.
n  f1  f 2  f 3  f 4  f 5  4  8  5  2  1  20
and
M1  12, M 2  17, M 3  22, M 4  27, M 5  32.
Thus,
5
xg 
fM
i
i 1
i

5
f
i 1
4  12  8  17   5  22  2  27   1 32
 19
4  8  5  2 1
i
and
 f M
5
s g2 
i 1
i
 xg 
2
i
n 1
2
2
2
2
2
4  12  19  8  17  19  5  22  19  2  27  19  1 32  19

20  1
 30
Online Exercise:
Exercise 3.5.1
4