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 :

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)

(2) - 3(1) :

-4f₂  =  -48

f₂  =  12

Substitute f₂  =  12 in (1).

f₁ + 12  =  20

f₁  =  8

So, 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 :

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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