Question

If mode =125 find x 100-120 120-140 140-160 160-180 180-200 Frequency- 12,14,x,6,10

Answer 1

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Answer 2

Given:

Mode = 125

$\begin{tabular}{| c | c |}\cline{1-2}&&\sf Class & \sf Frequency\\&&\cline{1-2}&&\sf 100 \ - \ 120 & \sf 12 \\&&\cline{1-2}&&\sf 120 \ - \ 140 & \sf 14 \\&&\cline{1-2}&&\sf 140 \ - \ 160 & \sf x \\&&\cline{1-2}&&\sf 160 \ - \ 180 & \sf 6 \\&&\cline{1-2}&&\sf 180 \ - \ 200 & \sf 10 \\&&\cline{1-2}\end{tabular}$

To find:

The value of x.

What is the mode of a given frequency distribution table?

The mode of the data is the most common value, and the class which belongs to the highest frequency is the Modal Class.

Solution:

In the given data, the modal frequency is 14.

The modal class is 120 - 140.

$\begin{tabular}{| c | c |}\cline{1-2}&&\sf Class & \sf Frequency \\&&\cline{1-2}&&\sf 100 \ - \ 120 & \sf 12 \\&&\cline{1-2}&&\sf 120 \ - \ 140 & \sf 14 \\&&\cline{1-2}&&\sf 140 \ - \ 160 & \sf x \\&&\cline{1-2}&&\sf 160 \ - \ 180 & \sf 6 \\&&\cline{1-2}&&\sf 180 \ - \ 200 & \sf 10 \\&&\cline{1-2}&&& \Sigma \sf Fi = 42 + x\\&&\cline{1-2}\end{tabular} \begin{tabular}{c}&\longmapsto \sf Modal \ Class \\ & \\&&&&&\end{tabular}$

We know that the mode of a data is given by;

$\Longrightarrow \sf Mode = \ell + \Bigg( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \Bigg) \times h$

Definition of terms used:

(Let us mention "Frequency" as F)

⇔ $\ell$ = Lower limit of the modal class = 120

⇔ f₀ = F of the class preceding the F of modal class = 12

⇔ f₁ = F of the modal class = 14

⇔ f₂ = F of the class succeeding the F of modal class = x

⇔ h = Size of the class interval. (Upper Limit - Lower Limit) = 20

$\Longrightarrow \sf Mode = 125$

$\Longrightarrow \sf 125 = \ell + \Bigg( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \Bigg) \times h$

$\Longrightarrow \sf 125 = 120 + \Bigg( \dfrac{14 - 12}{2(14) - 12 - x} \Bigg) \times 20$

$\Longrightarrow \sf 125 - 120 = \Bigg( \dfrac{2}{28 - 12 - x} \Bigg) \times 20$

$\Longrightarrow \sf 5 = \Bigg( \dfrac{2}{16 - x} \Bigg) \times 20$

$\Longrightarrow \sf 5 = \Bigg( \dfrac{40}{16 - x} \Bigg)$

$\Longrightarrow \sf 5\Big(16 - x\Big) =40$

$\Longrightarrow \sf 80 - 5x =40$

$\Longrightarrow \sf 80 - 40 = 5x$

$\Longrightarrow \sf 5x = 40$

$\Longrightarrow \sf x = 40/5$

$\Longrightarrow \sf x = 8$

Therefore, the value of 'x' is 8.