The mean of the following data is 53, Find the values of f1 and f2.
C.I : 0–20 20–40 40–60 60–80 80–100 .
f1 : 15 f1 21 f2 17 Total 100 .
Answers
The mean of a number of observation is the sum of the values of all the observations divided by the total number of observations.
Mean=sum of all the observations/total number of observations
Direct method:
In this method find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by xi. For each class interval we have the frequency( fi) corresponding to the class mark (xi).
CLASS MARKS =( lower limit + upper limit)/2
Then find the product of fi, & xi for each class interval. Find Σfi & Σfixi.
Use the formula :
Mean = Σfixi/ Σfi
SOLUTION IS IN THE ATTACHMENT.
HOPE THIS WILL HELP YOU....
search-icon-header
Search for questions & chapters
search-icon-image
Question
Bookmark
The mean of the followlng frequency table is 53. But the frequencies f
1
and f
2
in the classes 20−40 and 60−80 are missing. Find the missinq frequencies.
Age (in years)
0-20
20-40
40-60
60-80
80-100
Total
No. of people
15
f
1
21
f
2
17
100
Hard
Solution
verified
Verified by Toppr
Correct option is D)
Answer:-
Age No. of people
(f) x
i
=
2
lower limit + upper limit
x
i
f
i
0-20 15 10 150
20-40 f
1
30 30f
1
40-60 21 50 1050
60-80 f
2
70 70f
2
80-100 17 90 1530
Σf
i
=53+f
1
+f
2
Σf
i
.x
i
=2730+30f
1
+70f
2
Given:- Σf
i
=100
⇒53+f
1
+f
2
=100
⇒f
1
+f
2
=100−53
⇒f
1
+f
2
=47⟶eq.(i)
Given:- Mean =
Σf
i
Σx
i
.f
i
=53
⇒
100
2730+f
1
+f
2
=53
⇒273+f
1
+f
2
=530
f
1
+f
2
=257⟶eq.(ii)
from eq. (i) & (ii), we get
f
1
=18,f
2
=29
⇒f
1
=18,f
2
=29