If the mean of the following distribution is 2-6 then the value of y is begin array - WhlineVariablex-1-2-3-4-5Whline Frequency-4-5-y-1-2Whline array

The correct option is B 8n=∑ f_i=4+5+y+1+2=12+y ∑ f_ix_i=4+10+3y+4+10=28+3y Mean=∑ f_ix_i/n⇒ 2.6=28+3y/12+y ⇒ 26(12+y)=10(28+3y)⇒ y=8 ...

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Standard IX

Mathematics

Mean

If the mean o...

Question

If the mean of the following distribution is 2.6 then the value of y is
$\begin{matrix} V a r i a b l e (x) & 1 & 2 & 3 & 4 & 5 F r e q u e n c y & 4 & 5 & y & 1 & 2 \end{matrix}$

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Solution

The correct option is B 8
$n = \sum f_{i} = 4 + 5 + y + 1 + 2 = 12 + y$
$\sum f_{i} x_{i} = 4 + 10 + 3 y + 4 + 10 = 28 + 3 y$
$M e a n = \frac{\sum f_{i} x_{i}}{n} \Rightarrow 2.6 = \frac{28 + 3 y}{12 + y} \Rightarrow 26 (12 + y) = 10 (28 + 3 y) \Rightarrow y = 8$

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Q. If the mean of the following distribution is 2.6, then the value of y is

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Q. If the mean of the distribution is 2.6, then the value of y is:

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If the mean of the following distribution is 2.6, then the value of y is

Q. The mean of the above frequency distribution is 3.5, then find the value of x.

\begin{matrix} V a r i a b l e & 1 & 2 & x & 4 & 5 F r e q u e n c y & 2 & 3 & 4 & 5 & 6 \end{matrix}

The mean of the above frequency distribution is 3.5, then the value of x is:
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If the mean of the following distribution is 2.6 then the value of y is Variable(x)12345Frequency45y12

The correct option is B 8n=∑fi=4+5+y+1+2=12+y ∑fixi=4+10+3y+4+10=28+3y Mean=∑fixin⇒2.6=28+3y12+y⇒26(12+y)=10(28+3y)⇒y=8

If the mean of the following distribution is 2.6 then the value of y is
$\begin{matrix} V a r i a b l e (x) & 1 & 2 & 3 & 4 & 5 F r e q u e n c y & 4 & 5 & y & 1 & 2 \end{matrix}$

The correct option is B 8
$n = \sum f_{i} = 4 + 5 + y + 1 + 2 = 12 + y$
$\sum f_{i} x_{i} = 4 + 10 + 3 y + 4 + 10 = 28 + 3 y$
$M e a n = \frac{\sum f_{i} x_{i}}{n} \Rightarrow 2.6 = \frac{28 + 3 y}{12 + y} \Rightarrow 26 (12 + y) = 10 (28 + 3 y) \Rightarrow y = 8$