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Question

The mean of the following data is 42. Find the missing frequencies x and y if the sum of frequencies is 100.
Class interval0101020203030404050506060707080Frequency710x13y10149

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Solution

The given data is shown as follows:

Class interval Frequency (fi) Class mark (xi) (fixi)
0 - 10 7 5 35
10 - 20 10 15 150
20 - 30 x 25 25x
30 - 40 13 35 455
40 - 50 y 45 45y
50 - 60 10 55 550
60 - 70 14 65 910
70 - 80 9 75 675
Total ∑fi=63+x+y ∑(fi×xi)=2775+25x+45y

Sum of the frequencies = 100

⇒ ∑fi = 100

⇒ 63 + x + y = 100

⇒ x + y = 100 - 63

⇒ x + y = 37

⇒ y = 37 - x – – – – – – (1)

Now, the mean of the given data is given by

¯x = ∑(fi×xi)/∑fi

⇒42= 2775+25x+45y100

⇒ 4200 = 2775 + 25 x + 45y

⇒4200 - 2775 = 25x + 45y

⇒ 1425 = 25x + 45(37 - x) [from (1)]

⇒ 1425 = 25x + 1665 - 45x

⇒ 20x = 1665 - 1425

⇒ 20x = 240

⇒ x = 12

If x = 12, then y = 37 - 12 = 25

Thus, the value of x is 12 and y is 25.


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