Question

53. The Mode of given frequency distribution is 37. Findthe missing frequency x.
Class
0-10
10-20
20-30
30-40
40-50
50-60
60-70
frequency
4.
7.
х
18
15
8
7​

Answer 1

$\boxed{\begin{array}{c|c|c|c|c|c|c|c}\bf Class\: interval&\sf 0-10&\sf 10-20&\sf 20-30 &\sf 30-40&\sf 40-50&\sf 50-60&\sf 60-70\\\frac{\qquad \quad \qquad}{}&\frac{\quad \quad \quad}{}&\frac{\quad \quad\quad}{}&\frac{\quad \quad \quad}{}&\frac{\quad \quad \quad}{}&\frac{\quad \quad\quad}{}&\frac{\quad \quad\quad}{}&\frac{\quad \quad\quad}{}\\\bf Frequency&\sf 4&\sf 7&\sf x&\sf 18&\sf 15&\sf 8&\sf 7\end{array}}$

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Here, Maximum frequency is 18.

Then, the corresponding class 30-40 is the model class.

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$\dag\;{\underline{\frak{Formula\:of\:Mode\:is\:given\:by,}}}$

$\star\;{\boxed{\sf{\pink{Mode = l + \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h}}}}$

Here,

• $\sf Lower\:limit\:of\;class,\:l = \bf{30}$
• $\sf Class\:interval,\:h = 40 - 30 = \bf{10}$
• $\sf Frequency\:of\:Median\;class\:,f_1 = \bf{18}$
• $\sf Frequency\:of\:preceding\;class\:,f_0 = \bf{x}$
• $\sf Frequency\:of\: succeeding\;class\:,f_2 = \bf{15}$

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$\dag\;{\underline{\frak{Putting\:values\:in\:formula,}}}\\ \\ \\:\implies\sf 37 = 30 + \dfrac{18 - x}{2 \times 18 - x - 15} \times 10\\ \\ \\:\implies\sf 37 - 30 =\dfrac{18 - x}{36 - x - 15} \times 10\\ \\ \\:\implies\sf 7 = \dfrac{18 - x}{21 - x} \times 10\\ \\ \\:\implies\sf 7(21 - x) =10(18 - x)\\ \\ \\:\implies\sf 147 - 7x = 180 - 10x\\ \\ \\ :\implies\sf - 7x + 10x = 180 - 147\\ \\ \\:\implies\sf 3x = 33\\ \\ \\:\implies\sf x = \cancel{\dfrac{33}{3}}\\ \\ \\ :\implies{\underline{\boxed{\pmb{\frak{\purple{x = 11}}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{The\:value\:of\:Missing\:Frequency\:(x)\:is\: {\textsf{\textbf{11}}}.}}}$

Answer 2

Answer:

Given :-

Class Interval - 0-10, 10-20, 20-30, 30-40, 40-50,50-60, 60-70

Frequency - 4,7,x,18,15,8,7

Mode = 37

To Find :-

Value of x

Solution :-

We know that

$\bf Mode = l +\bigg( \dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg)\times h$

We have

$\sf\begin{cases} \sf l = 30 \\ \sf h = 10 \\ \sf f_1 = 18 \\ \sf f_2 = 15\\ \sf f_0 = x\end{cases}$

$\sf \implies \: 30 + \dfrac{18 - x}{2(18) - x - 15} \times 10 = 37$

$\sf \implies \: 30 + \dfrac{18 - x}{36 - x - 15} \times 10 = 37$

$\sf \implies \: \dfrac{18 - x}{21 - x} \times 10 = 37 - 30$

$\sf \implies \: \dfrac{18 - x}{21 - x} \times 10 = 7$

$\sf \implies \: 10(18 - x) = 7(21 - x)$

$\sf \implies \: 180 - 10x = 147 - 7x$

$\sf \implies \: 180 - 147 = 10x - 7x$

$\sf \implies \: 33 = 3x$

$\sf \implies \: x = \dfrac{33}{3}$

$\bf \red{x = 11}$