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The frequency distribution of life
$90 \mathrm{TV}$ tubes whose median life is $17 \mathrm{months}$ is as follows
\begin{array}{|c|c|}
\hline Life of TV tubes (in months) & No. of TV tubes \\
\hline 0-5 & 3 \\
\hline 5-10 & 12 \\
\hline 10-15 & x \\
\hline 15-20 & 35 \\
\hline 20-25 & y \\
\hline 25-30 & 4 \\
\hline
\end{array} $\mathrm{n}=90$
$\frac{\mathrm{n}}{2}=45$
\begin{array}{|c|c|c|}
\hline Calss & Frequency & c f \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
What is the lower limit of the median class? $\quad$
Options:
$90 \mathrm{TV}$ tubes whose median life is $17 \mathrm{months}$ is as follows
\begin{array}{|c|c|}
\hline Life of TV tubes (in months) & No. of TV tubes \\
\hline 0-5 & 3 \\
\hline 5-10 & 12 \\
\hline 10-15 & x \\
\hline 15-20 & 35 \\
\hline 20-25 & y \\
\hline 25-30 & 4 \\
\hline
\end{array} $\mathrm{n}=90$
$\frac{\mathrm{n}}{2}=45$
\begin{array}{|c|c|c|}
\hline Calss & Frequency & c f \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
What is the lower limit of the median class? $\quad$
Solution:
1669 Upvotes
Verified Answer
The correct answer is:
15
\begin{array}{|c|c|c|}
\hline Class & Frequency & C. \mathbf{F} \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
Since, $n=90 \therefore \frac{n}{2}=45$
$\because$ Lower limit of median class is 15 .
\hline Class & Frequency & C. \mathbf{F} \\
\hline 0-5 & 3 & 3 \\
\hline 5-10 & 12 & 15 \\
\hline 10-15 & x & 15+x \\
\hline 15-20 & 35 & 50+x \\
\hline 20-25 & y & 50+x+y \\
\hline 25-30 & 4 & 54+x+y \\
\hline
\end{array}
Since, $n=90 \therefore \frac{n}{2}=45$
$\because$ Lower limit of median class is 15 .
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