Find the mode

$\begin{matrix} C l a s s i n t e r v a l & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 F r e q u e n c y & 8 & 9 & 11 & 12 & 10 \end{matrix}$

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Solution

Modal Class is 130-140

Mode=I+ $(\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}})$ $\times$ h

= 130 + $\frac{12 - 11}{24 - 11 - 10}$ $\times$ 10=133.33

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Similar questions

Find the mode.

$\begin{matrix} C l a s s i n t e r v a l & F r e q u e n c y 100 - 110 & 8 110 - 120 & 9 120 - 130 & 11 130 - 140 & 12 140 - 150 & 10 \end{matrix}$ [2 MARKS]

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$\begin{matrix} C l a s s i n t e r v a l & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 F r e q u e n c y & 6 & 9 & 14 & 2 & 19 \end{matrix}$

Q. Find the mode.

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Find the mean of the following distribution using assumed mean.

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Find the median of the following data.

$\begin{matrix} C l a s s i n t e r v a l & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 F r e q u e n c y & 6 & 9 & 14 & 2 & 19 \end{matrix}$

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Find the mode Class interval100−110110−120120−130130−140140−150Frequency89111210 ___

Modal Class is 130-140 Mode=I+(f1−f02f1−f0−f2) ×h = 130 +12−1124−11−10×10=133.33

Find the mode

$\begin{matrix} C l a s s i n t e r v a l & 100 - 110 & 110 - 120 & 120 - 130 & 130 - 140 & 140 - 150 F r e q u e n c y & 8 & 9 & 11 & 12 & 10 \end{matrix}$

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Modal Class is 130-140

Mode=I+ $(\frac{f_{1} - f_{0}}{2 f_{1} - f_{0} - f_{2}})$ $\times$ h

= 130 + $\frac{12 - 11}{24 - 11 - 10}$ $\times$ 10=133.33