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If the mean of the following frequency distribution is 91 and $x - y = 14$ , find the missing frequencies.

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36 and 22

34 and 20

44 and 25

34 and 22

answer is B.

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Detailed Solution

Given that the mean of the given observations is 91.
Given frequency distribution table is:
Question Image

We need to find the value of x and y.
The formula of the mean is

\frac{\sum_{}^{} f_{i} x_{i}}{\sum_{}^{} f_{i}} .

In the above formula

\sum_{}^{} f_{i} x_{i}

is the sum of products of frequencies and their respective mid-interval of each class.

\Rightarrow \sum_{}^{} f_{i} x_{i} = (12 \times 15) + (21 \times 45) + (x \times 75) + (52 \times 105) + (y \times 135) + (11 \times 165)

\Rightarrow \sum_{}^{} f_{i} x_{i} = 180 + 945 + 75 x + 5460 + 135 y + 1815

\Rightarrow \sum_{}^{} f_{i} x_{i} = 8400 + 75 x + 135 y

Now, summation of frequencies is,

\Rightarrow \sum_{}^{} f_{i} = 12 + 21 + x + 52 + y + 11

\Rightarrow \sum_{}^{} f_{i} = 96 + x + y

Now, mean of the given data is:

\Rightarrow mean = \frac{8400 + 75 x + 135 y}{96 + x + y}

\Rightarrow 91 = \frac{8400 + 75 x + 135 y}{96 + x + y}

\Rightarrow 91 (96 + x + y) = 8400 + 75 x + 135 y

\Rightarrow 8736 - 8400 + 91 + 91 y = 75 x + 135 y

\Rightarrow (75 - 91) x + (135 - 91) y = 336

\Rightarrow - 16 x + 44 y = 336 . . . . . (i)

As,

x - y = 14 (given)

\Rightarrow x = y + 14

By substituting value of x in (i), we get

\Rightarrow - 16 (y + 14) + 44 y = 336

\Rightarrow - 16 y - 224 + 44 y = 336

\Rightarrow - 16 y + 44 y = 336 + 224

\Rightarrow 28 y = 560

\Rightarrow y = 20

As y = 20, then

\Rightarrow x = 20 + 14

\Rightarrow x = 34

Therefore, the values of x and y are 34 and 20.
Hence, option 2 is the correct answer.

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