Consider the frequency distribution of the given numberFrequency-5-4-6-fillhline lend array If the mean is known to be 3 - then the value of f isA- 10B- 15C- 14D- 3

The correct option is C 14Mean=1×5+2×4+3×6+4× f/15+f i.e., 3=5+8+18+4f/15+f⇒ 45+3f=31+4f ⇒ 45 31=f⇒ f=14 ...

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Standard IX

Mathematics

Approach to Find Mean for Frequency Distribution Table

Consider the ...

Question

Consider the frequency distribution of the given number
$\begin{matrix} V a l u e : & 1 & 2 & 3 & 4 F r e q u e n c y : & 5 & 4 & 6 & f \end{matrix}$
If the mean is known to be 3, then the value of $f$ is ____.

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Solution

The correct option is C 14
$M e a n = \frac{1 \times 5 + 2 \times 4 + 3 \times 6 + 4 \times f}{15 + f}$
$i . e ., 3 = \frac{5 + 8 + 18 + 4 f}{15 + f} \Rightarrow 45 + 3 f = 31 + 4 f$
$\Rightarrow 45 - 31 = f \Rightarrow f = 14$

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Consider the frequency distribution of the given number Value:1234Frequency:546f ​If the mean is known to be 3, then the value of f is ____.

The correct option is C 14Mean=1×5+2×4+3×6+4×f15+f i.e., 3=5+8+18+4f15+f⇒45+3f=31+4f ⇒45−31=f⇒f=14

Consider the frequency distribution of the given number
$\begin{matrix} V a l u e : & 1 & 2 & 3 & 4 F r e q u e n c y : & 5 & 4 & 6 & f \end{matrix}$
If the mean is known to be 3, then the value of $f$ is ____.

The correct option is C 14
$M e a n = \frac{1 \times 5 + 2 \times 4 + 3 \times 6 + 4 \times f}{15 + f}$
$i . e ., 3 = \frac{5 + 8 + 18 + 4 f}{15 + f} \Rightarrow 45 + 3 f = 31 + 4 f$
$\Rightarrow 45 - 31 = f \Rightarrow f = 14$