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 students whose marks are less than 12 is 18 , then the total number of students is :

<p>The median for grouped data is given by:</p> <p>$\text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h$</p> <p>where</p> <p><p>$L$ is the lower limit (or boundary) of the median class.</p></p> <p><p>$CF$ is the cumulative frequency of all classes preceding the median class.</p></p> <p><p>$f$ is the frequency of the median class.</p></p> <p><p>$h$ is the class width.</p></p> <p><p>$n$ is the total number of students.</p></p> <p>Given:</p> <p><p>Median $= 14$</p></p> <p><p>Median class interval is $12-18$, so $L = 12$ and the class width $h = 18 - 12 = 6$.</p></p> <p><p>Frequency of median class $f = 12$.</p></p> <p><p>Cumulative frequency below the median class $= 18$ (i.e., $CF = 18$).</p></p> <p>Plugging these into the formula:</p> <p>$14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6$</p> <p>Step 1: Subtract 12 from both sides:</p> <p>$2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6$</p> <p>Step 2: Simplify the multiplication factor:</p> <p>$\left(\frac{6}{12}\right) = \frac{1}{2}$</p> <p>So the equation becomes:</p> <p>$2 = \frac{1}{2}\left(\frac{n}{2} - 18\right)$</p> <p>Step 3: Multiply both sides by 2 to remove the fraction:</p> <p>$4 = \frac{n}{2} - 18$</p> <p>Step 4: Solve for $\frac{n}{2}$:</p> <p>$\frac{n}{2} = 4 + 18 = 22$</p> <p>Step 5: Multiply both sides by 2 to find $n$:</p> <p>$n = 44$</p> <p>Thus, the total number of students is $44$.</p>