Question

The mean of the following frequency distribution of 100 observations is 148. Find the missing frequencies f₁ and f₂
Class
0-49
50-99
100-149
150-199
200-249
250-299
300-349

Frequency
10
15
f₁
20
15
f₂
2

Answer 1

Dear student,

Answer: f1 = 30 , f2 = 8

Solution:

According to question total observation are 100.

So, summation of all frequencies would be 100.

62 + f1 + f2 = 100

f1 + f2 = 100-62

f1 + f2 = 38 --------------eq1

Mean = 148

148 = a + (∑( fi ui)/∑ fi )×h

From attached figure you get that a = 174.5 and h = 50

148 = 174.5 + (-39-f1-2f2 )/( 62+f1+f2 )× 50

(148 -174.5 )÷50 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

-53/100 = (-39-f1+2f2 ) ÷ (62+f1+f2 )

(62+f1+f2 )×(-53) = 100 ×(-39-f1+2f2 )

-3286 -53f1 -53f2 = -3900 -100 f1 +200f2

47 f1 -253 f2 = -614 ------eq2

multiply eq 1 by 47 and subtract both equation

47f1 - 47 f1 + 253 f2 + 47 f2 = 1786 +614

300 f2 = 2400

f2 = 2400/300

f2 = 8

put the value of f2 in equation 1

f1+f2 = 38

f1 = 38 -8

f1 = 30

Hope it helps you.

Answer 2

SOLUTION IS IN THE ATTACHMENT..
ANSWER :
THE MISSING FREQUENCIES ARE f1 = 30 & f2 = 8.
MEAN:
The mean of a number of observation is the sum of the values of all the observations divided by the total number of observations.
To calculate the mean of grouped data we have three methods which are:
i)Direct method
(ii)The assumed mean
(iii)The Step deviation
STEP DEVIATION METHOD:
Step deviation method is used in the cases where the deviation from the assumed mean multiples of a common number. If the values of di for each class is a multiple of h the calculation become simpler by taking ui= di/h = (xi - A )/h
Here, h is the class size of each class interval.
Mean = A +( Σfiui/Σfi) × h
HOPE THIS ANSWER WILL HELP YOU...