The following is a frequency distribution for the ages of a sample of employees
(a) Determine the average age for the sample.
(b) Compute the standard deviation.
Ther
more dispersed distribution.
local company
at a
ople.
Age (in years)
Frequency
30 – 39
ghts of 50
2
40 - 49
3
50 - 59
7
60 - 69
5
70 - 79
1
y mean Em
(c) Compute the coefficient of variation.
Answers
Answer:
Total number of elements(n) =4+16+19+28+22+8+3=100
So, median group will be group of 50
th
and 51
th
elements.
Which is 40−45 age interval.
Now, median =L+
f
m
2
n
−cf
b
×w
Where,
L= Lower class boundary of the modal group,
f
m
= Frequency of the median group
n= Total number of the median group
cf
b
= Cumulative frequency of the groups before the median class
w= Group width.
Median =40+
38
50−(4+16+19)
×5
Median =40+1.96≡42
Hence, median is 42 years.
MEDIAN = 42
GIVEN: Frequency distribution for the ages of a sample of employees
TO FIND: Average Age/Median
SOLUTION:
As we know,
The average of data provided is called the median of the same.
It is calculated by adding up all the elements of data and further dividing it by the total number of elements.
In this question,
Number of elements = 7
Sum of all elements(n) =4+16+19+28+22+8+3
=100
So,
median group = group of 50th and 51st elements.
That is the 40−45 age interval.
Now, median =L+fm/2n−cfb×w
Where,
L= Lower class of the modal group,
fm= Frequency of the median group
n= Total number of the median group
cfb= Cumulative frequency of the groups before the median class
w= Group width.
Therefore,
Median =40+38/50−(4+16+19)×5
Median =40+1.96≡42
Hence, the median is 42 years.
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