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Mean

The mean of f...

Question

The mean of following frequency table is $50.$
If the total of frequency is $120.$ Then the missing frequencies $f_{1}, f_{2}$ are respectively
$\begin{matrix} Class & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 Frequency & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}$

24, 28

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28, 24

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25, 27

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27, 25

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Solution

The correct option is B $28, 24$
Given : $\begin{matrix} Class & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 Frequency & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}$

Total frequency $= 120$ , then
$17 + f_{1} + 32 + f_{2} + 19 = 120 \Rightarrow f_{1} + f_{2} = 52 \dots (1)$
We know the mean
$¯ ¯ ¯ x = \sum \frac{f_{i} x_{i}}{N} x_{1} = \frac{x_{i} + x_{f}}{2} = \frac{0 + 20}{2} = 10 x_{2} = \frac{x_{i} + x_{f}}{2} = \frac{20 + 40}{2} = 30 x_{3} = \frac{x_{i} + x_{f}}{2} = \frac{40 + 60}{2} = 50 x_{4} = \frac{x_{i} + x_{f}}{2} = \frac{60 + 80}{2} = 70 x_{5} = \frac{x_{i} + x_{f}}{2} = \frac{80 + 100}{2} = 90$
Therefore,
$¯ ¯ ¯ x = \frac{f_{1} x_{1} + f_{2} x_{2} + f_{3} x_{3} + f_{4} x_{4} + f_{5} x_{5}}{N} \Rightarrow 50 = \frac{17 \times 10 + f_{1} \times 30 + 32 \times 50 + f_{2} \times 70 + 19 \times 90}{120} \Rightarrow 6000 = 170 + 30 f_{1} + 1600 + 70 f_{2} + 1710 \Rightarrow 6000 = 30 f_{1} + 70 f_{2} + 3480 \Rightarrow 2520 = 30 f_{1} + 70 f_{2} \Rightarrow 3 f_{1} + 7 f_{2} = 252 \dots (2)$

Solving equation s $(1)$ and $(2)$ , we get
$f_{1} = 28, f_{2} = 24$

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Question 8
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Mean

Standard XI Mathematics

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The mean of following frequency table is 50. If the total of frequency is 120. Then the missing frequencies f1,f2 are respectively Class0−2020−4040−6060−8080−100Frequency17f132f219

The mean of following frequency table is $50.$
If the total of frequency is $120.$ Then the missing frequencies $f_{1}, f_{2}$ are respectively
$\begin{matrix} Class & 0 - 20 & 20 - 40 & 40 - 60 & 60 - 80 & 80 - 100 Frequency & 17 & f_{1} & 32 & f_{2} & 19 \end{matrix}$