Question

The mean of the following frequency distribution is 57.6 and the sum of the observations is 50.

Class: 0-20 20-40 40-60 60-80 80-100 100-120

Frequency: 7, f1, 12, f2, 8, 5

Find f1 and f2.

Answer 1

★ STATISTICS ★

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Explanation :-

The mean of a number of observation is the sum of the values of all the observations divided by the total number of observations.

So, $Mean= \frac{Sum of all the Observations}{Total Number of Observations}$

Now, Find the class marks of class interval. These class marks would serve as the representative of whole class and are represented by (fx). For each class interval we have the frequency (f) corresponding to the class mark (x).

From, $Class Marks = \frac{Lower Limit + Upper Limit}{2}$

Then, find the product of (f) & (x) for each class interval. Find (Σf) & (Σfx),

By, x̄ = Σfx / Σf

Calculations :-

Sum of the frequencies (∑f) $= 7 +f_{1} + 12 +f_{2} + 8 + 5 = 32 +f_{1} +f_{2}$

⇒ $32 +f_{1} +f_{2} = 50$
⇒ $f_{1} +f_{2} = 18$   ----- (1)

Now, Mean of the data :

Mid values (x) of the classes are 10, 30, 50, 70, 90, 110.
So, Product of the frequencies and mid-values,
 $(fx) = 70, 30f_{1} , 600, 70f_{2} , 720, 550$

⇒ ∑fx $= 70 + 30f_{1} + 600 + 70f_{2} + 720 + 550$

Thus, x̄ = ∑fx / ∑f 
      $= \frac{70 + 30f_{1} + 600 + 70f_{2} + 720 + 550}{50}$
      $= 57.6$

Also, $30 f_{1} + 70 f_{2} = 940$
⇒ $3 f_{1} + 7 f_{2} = 94$  ----- (2)
Solving equations (1) and (2), 

We get, $f_{1} = 8$ and $f_{2} = 10$

∴ The missing frequencies are 8 and 10.

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Regards

#ExoticExplorer