Given below is a frequency distribution with median 46 . In this...

Given below is a frequency distribution with median 46 . In this distribution, some of the frequencies are missing : Determine the missing frequencies.

<table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td>Marks</td><td>10-20</td><td>20-30</td><td>30-40</td><td>40-50</td><td>50-60</td><td>60-70</td><td>70-80</td><td>Total</td></tr><tr><td>No. of students</td><td>12</td><td>30</td><td>?</td><td>65</td><td>?</td><td>25</td><td>18</td><td>229</td></tr></table>

)

Options

A34,45
B25,40
C12,18
D30,35

Correct answer

A. 34,45

Step-by-step solution

$\begin{array}{|c|c|c|} \hline\text{ Marks }& \text{ No. of s tudents }& \text{ c.f. }\\ \hline 10-20 & 12 & 12 \\ \hline 20-30 & 30 & 42 \\ \hline 30-40 & ? & 42+\mathrm{x}\\ \hline 40-50 & 65 & 107+\mathrm{x}\\ \hline 50-60 & ? & 107+\mathrm{x}+\mathrm{y}\\ \hline 60-70 & 25 & 132+\mathrm{x}+\mathrm{y}\\ \hline 70-80 & 18 & 150+\mathrm{x}+\mathrm{y}\\ \hline \text{ Total }& 229 & \\ \hline \end{array}

)

\begin{aligned}& \frac{\mathrm{n}}{2}=\frac{229}{2}=114.5, \text{ Median }=46 \\ & \therefore \text{ Median class }=40-50 \\ & \therefore \ell=40, \text{ c.f. }=42+\mathrm{x}, \mathrm{f}=65, \mathrm{ h}=10 \\ & \text{ Median }=\ell+\left(\frac{\frac{\mathrm{n}}{2}-\text{ c.f. }}{\mathrm{f}}\right) \times \mathrm{h}\\ & 46=40+\frac{114.5-(42+\mathrm{x})}{65} \times10 \\ & \text{ or } \quad46-40=\frac{(114.5-42-\mathrm{x})}{13} \times2 \\ & 6=\frac{(72.5-\mathrm{x})}{13} \times2 \text{ or }78=145-2 \mathrm{x}\\ & 2 \mathrm{x}=145-78=67 \quad \text{ or } \mathrm{x}=\frac{67}{2}=33.5 \end{aligned}

) (\therefore \mathrm{x}=34(\because) Number of students cannot be in fraction) Now (\Sigma f_i=29 \therefore \mathrm{x}+\mathrm{y}+150=229)

x+y=229-150=79

...(i) Putting the value of

x$ in (i), we get (34 x+y=79 \Rightarrow y=79-34=45) (\therefore \quad \mathrm{x}=34, \mathrm{y}=45)

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