Marks obtained by all the students of class 12 are presented in a frequency 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:
Solutions · 1
Official solution
The median for grouped data is given by:
\[ \text{Median} = L + \left(\frac{\frac{n}{2} - CF}{f}\right) \times h \]
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:
\[ 14 = 12 + \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 \]
Subtract 12 from both sides: 1: Subtract 12 from both sides:
\[ 2 = \left(\frac{\frac{n}{2} - 18}{12}\right) \times 6 \]
Simplify the multiplication factor: 2: Simplify the multiplication factor:
\[ \left(\frac{6}{12}\right) = \frac{1}{2} \]
So the equation becomes:
\[ 2 = \frac{1}{2}\left(\frac{n}{2} - 18\right) \]
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.