# FINDING MISSING FREQUENCY WHEN MEAN IS GIVEN

Finding Missing Frequency When Mean is Given :

Here we are going to see, some example problems to find the missing frequency when mean is given.

## Finding Missing Frequency When Mean is Given

Question 1 :

The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2.

Solution :

 x0 - 2020 - 4040 - 6060 - 8080 - 100100 - 120 midpoint (x)1030507090110 f5f110f278 fx5030f150070f2630880

Mean  =  62.8

Σf  =  5 + f1 + 10 + f2 + 7 + 8

Σf  =  30 + f1 + f2

50  =  30 + f1 + f2

f1 + f2  =  20 ---(1)

Σfx  =  50 + 30f1 + 500 + 70f2 + 630 + 880

Σfx  =  2060 + 30f1 + 70f2

Σfx/Σf  =  (2060 + 30f1 + 70f2 )/50  =  62.8

2060 + 30f1 + 70f2  =  62.8(50)

2060 + 30f1 + 70f2  =  3140

30f1 + 70f2  =  3140 - 2060

30f1 + 70f2  =  1080

3f1 + 7f2  =  108  ----(2)

(1) 3  => 3f1 + 3f2  =  60

3f1 + 7f2  =  108

(-)    (-)       (-)

-----------------------

-4f2  =  -48  ==> f2  =  12

By applying the value of f2 in (1), we get

f1 + 12  =  20

f1  =  20 - 12  =  8

Hence the missing frequencies are 8 and 12.

Question 2 :

The diameter of circles (in mm) drawn in a design are given below

Calculate the standard deviation.

Solution :

 32.5 - 36.536.5 - 41.540.5 - 44.544.5 - 48.548.5 - 52.5 mid34.53942.546.550.5 f1517212225 d = x - A-8, 64-3.5, 12.250, 04, 168, 64 fd-120-59.5088200

Σfd   =  -120 - 59.5 + 0 + 88 + 200

Σfd  =  108.5

Σfd/Σf  =  108.5/100  =  1.085

Σfd2  =   15(64) + 17(12.25) + 21(0) + 22(16) + 25(64)

=  960 + 208.25 + 0 + 352 + 1600

=  3120.25

Σfd2/Σf  =  3120.25/100

=  31.2025

N = Σf = 100

Standard deviation  =  √31.20 - (1.085)2

=  √31.20 -  1.18

= 5.47 (APPROXIMATELY)

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