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The frequency distribution of the age of students in a class of 40 students is given below.
$\begin{array}{|l|c|c|c|c|l|l|}
\hline \text{Age} & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline \text{No of Students} & 5 & 8 & 5 & 12 & x & y \\
\hline
\end{array}$
If the mean deviation about the median is 1.25, then $4 x+5 y$ is equal to :
Options:
$\begin{array}{|l|c|c|c|c|l|l|}
\hline \text{Age} & 15 & 16 & 17 & 18 & 19 & 20 \\
\hline \text{No of Students} & 5 & 8 & 5 & 12 & x & y \\
\hline
\end{array}$
If the mean deviation about the median is 1.25, then $4 x+5 y$ is equal to :
Solution:
1358 Upvotes
Verified Answer
The correct answer is:
44
$\begin{aligned} & \mathrm{x}+\mathrm{y}=10 \ldots \ldots \ldots .(1) \\ & \text {Median }=18=\mathrm{M} \\ & \text {M.D. }=\frac{\sum \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right|}{\sum \mathrm{f}_{\mathrm{i}}} \\ & 1.25=\frac{36+\mathrm{x}+2 \mathrm{y}}{40} \\ & \mathrm{x}+2 \mathrm{y}=14 \ldots \ldots \ldots .(2) \\ & \text { by (1) & (2) } \\ & \mathrm{x}=6, \mathrm{y}=4 \\ & \Rightarrow 4 \mathrm{x}+5 \mathrm{y}=24+20=44\end{aligned}$
$\begin{array}{|l|l|l|l|}
\hline \operatorname{Age}\left(\mathrm{x}_{\mathrm{i}}\right) & \mathrm{f} & \left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| & \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| \\
\hline 15 & 5 & 3 & 15 \\
\hline 16 & 8 & 2 & 16 \\
\hline 17 & 5 & 1 & 5 \\
\hline 18 & 12 & 0 & 0 \\
\hline 19 & \mathrm{x} & 1 & \mathrm{x} \\
\hline 20 & \mathrm{y} & 2 & 2 \mathrm{y} \\
\hline
\end{array}$
$\begin{array}{|l|l|l|l|}
\hline \operatorname{Age}\left(\mathrm{x}_{\mathrm{i}}\right) & \mathrm{f} & \left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| & \mathrm{f}_{\mathrm{i}}\left|\mathrm{x}_{\mathrm{i}}-\mathrm{M}\right| \\
\hline 15 & 5 & 3 & 15 \\
\hline 16 & 8 & 2 & 16 \\
\hline 17 & 5 & 1 & 5 \\
\hline 18 & 12 & 0 & 0 \\
\hline 19 & \mathrm{x} & 1 & \mathrm{x} \\
\hline 20 & \mathrm{y} & 2 & 2 \mathrm{y} \\
\hline
\end{array}$
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