Marks obtains by all the students of class 12 are presented in a freqency distribution wit…

Question

Marks obtains by all the students of class 12 are presented in a freqency distribution with classes of equal width. Let the median of this grouped data be 14 with median class interval 12-18 and median class frequency 12. If the number of s...

Accepted Answer

The median for grouped data is given by:

$=L+(\frac{\frac{n}{2}-CF}{f})	imes h$

where

$L$ is the lower limit (or boundary) of the median class.

$CF$ is the cumulative frequency of all classes preceding the median class.

$f$ is the frequency of the median class.

$h$ is the class width.

$n$ is the total number of students.

Given:

Median $=14$

Median class interval is $12-18$, so $L=12$ and the class width $h=18-12=6$.

Frequency of median class $f=12$.

Cumulative frequency below the median class $=18$ (i.e., $CF=18$).

Plugging these into the formula:

$14=12+(\frac{\frac{n}{2}-18}{12})	imes 6$

Subtract 12 from both sides: 1: Subtract 12 from both sides:

$2=(\frac{\frac{n}{2}-18}{12})	imes 6$

Simplify the multiplication factor: 2: Simplify the multiplication factor:

$(\frac{6}{12})=\frac{1}{2}$

So the equation becomes:

$2=\frac{1}{2}(\frac{n}{2}-18)$

Multiply both sides by 2 to remove the fraction: 3: Multiply both sides by 2 to remove the fraction:

$4=\frac{n}{2}-18$

Solve for $\frac{n}{2}$: 4: Solve for $\frac{n}{2}$:

$\frac{n}{2}=4+18=22$

Multiply both sides by 2 to find $n$: 5: Multiply both sides by 2 to find $n$:

$n=44$

Thus, the total number of students is $44$.

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