Question

The mean of the following frequency distribition is 62.8 and sum of all frequencies is 50 .compute the missing frequencies of f1 and f2
Class  0-20  20-40  40-60  60-80  80-100 100-120
Frequency  5  f1  10  f2  7  8

Answer 1

Therefore
f1=8,
f2=12
i hope this will useful to u

Answer 2

Answer:

The value of f1 is 8 and f2 is 12.

Solution:

Let us construct a table to represent the given data:

To compute the mean, let us create a frequency distribution table as follows:

Given,

$\begin{array} { c } { \sum f _ { 1 } = 50 } \\\\ { 30 + f 1 + f 2 = 50 } \\\\ { f 1 + f 2 = 50 - 30 } \\\\ { f 1 + f 2 = 20 } \\\\ { f 1 = 20 - f 2 \ldots \ldots ( i ) } \end{array}$

Given  

Mean = 62.8

$\begin{array} { l } { \sum f _ { 1 } \mathrm { x } _ { \mathrm { i } } / \sum f _ { 1 } = 62.8 } \\\\ { \frac { 2060 + 30 \mathrm { fl } + 70 \mathrm { f } 2 } { 50 } = 62.8 } \\\\ { 2060 + 30 \mathrm { f } 1 + 70 \mathrm { f } 2 = 62.8 \times 50 } \\\\ { 2060 + 30 \mathrm { f } 1 + 70 \mathrm { f } 2 = 3140 } \\\\ { 30 \mathrm { f } 1 + 70 \mathrm { f } 2 = 3140 - 2060 } \end{array}$

$\begin{array} { l } { 30 \mathrm { f } 1 + 70 \mathrm { f } 2 = 1080 } \\\\ { 10 ( 3 \mathrm { f } 1 + 7 \mathrm { f } 2 ) = 1080 } \\\\ { 3 \mathrm { f } 1 + 7 \mathrm { f } 2 = \frac { 1080 } { 10 } = 108 } \end{array}$

From (i), putting f1 = 20 – f2 in the above equation, we get,

$\begin{array} { l } { 3 ( 20 - f 2 ) + 7 f 2 = 108 } \\\\ { 60 - 3 f 2 + 7 f 2 = 108 } \\\\ { 4 f 2 = 108 - 60 = 48 } \\\\ { f 2 = \frac { 48 } { 4 } = 12 } \\\\ { f 1 = 20 - 12 = 8 } \end{array}$

Thus, the value of f1 is 8 and f2 is 12.