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

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

⠀⠀⠀⠀

Here, Maximum frequency is 18.

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

⠀⠀⠀⠀

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

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

Here,

$\sf Lower\:limit\:of\;class,\:l = \bf{30}$

$\sf Class\:interval,\:h = 40 - 30 = \bf{10}$

$\sf Frequency\:of\:Median\;class\:,f_1 = \bf{18}$

$\sf Frequency\:of\:preceding\;class\:,f_0 = \bf{x}$

$\sf Frequency\:of\: succeeding\;class\:,f_2 = \bf{15}$

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

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

Answer 2

Answer:

⠀⠀⠀⠀

Here, Maximum frequency is 18.

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

⠀⠀⠀⠀

\begin{gathered}\begin{gathered}\dag\;{\underline{\frak{Formula\:of\:Mode\:is\:given\:by,}}}\\ \\\end{gathered}\end{gathered}

†

FormulaofModeisgivenby,

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

⋆

Mode=l+

2f

1

−f

0

−f

2

f

1

−f

0

×h

Here,

\sf Lower\:limit\:of\;class,\:l = \bf{30}Lowerlimitofclass,l=30

\sf Class\:interval,\:h = 40 - 30 = \bf{10}Classinterval,h=40−30=10

\sf Frequency\:of\:Median\;class\:,f_1 = \bf{18}FrequencyofMedianclass,f

1

=18

\sf Frequency\:of\:preceding\;class\:,f_0 = \bf{x}Frequencyofprecedingclass,f

0

=x

\sf Frequency\:of\: succeeding\;class\:,f_2 = \bf{15}Frequencyofsucceedingclass,f

2

=15

\begin{gathered}\begin{gathered}\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\\ \\\end{gathered}\end{gathered}

†

Puttingvaluesinformula,

:⟹37=30+

2×18−x−15

18−x

×10

:⟹37−30=

36−x−15

18−x

×10

:⟹7=

21−x

18−x

×10

:⟹7(21−x)=10(18−x)

:⟹147−7x=180−10x

:⟹−7x+10x=180−147

:⟹3x=33

:⟹x=

3

33

:⟹

x=11

x=11

★

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

ThevalueofMissingFrequency(x)is11.