(B) The formula for the median of grouped data is $\text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$. Given: $\text{Median} =

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(B) The formula for the median of grouped data is $\text{Median} = \ell + \left( \frac{\frac{N}{2} - F}{f} \right) \times h$. Given: $\text{Median} = 14$,$\ell = 12$,$h = 6$,$f = 12$,and $F = 18$. Substituting these values into the formula: $14 = 12 + \left( \frac{\frac{N}{2} - 18}{12} \right) \times 6$ $14 - 12 = \left( \frac{\frac{N}{2} - 18}{2} \right)$ $2 = \frac{\frac{N}{2} - 18}{2}$ $4 = \frac{N}{2} - 18$ $22 = \frac{N}{2}$ $N = 44$. Thus,the total number of students is $44$.

Class 11 Mathematics Statistics Word problem -Statistics

Medium

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

JEE MAIN 2025 All JEE MAIN

A
$48$
B
$44$
C
$40$
D
$52$

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The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ respectively. Later on,it was found that one of the observations was taken as $50$ in place of $40$. If the wrong entry is replaced by the correct one,then the sum of the squares of all the observations is:

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If the mean of the following frequency distribution is $2.6$,find the value of $f$.

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