Question

The following data gives the marks obtained by 50 students in a test:

Marks	Number of Students
0-10	8
10-20	4
20-30	10
30-40	12
40-50	13
50-60	3

If $60\%$ of the students pass the test, find the minimum marks obtained by a pass student.

Answer 1

We can find the cumulative frequency of the class. Then we can find at which position the lowest pass mark lies. Then we can find the required class and find the minimum marks by giving substitution to formula to find the median after changing the half of the observations to 40 percent of the observation.

Complete step by step solution:
We are given the data of the marks obtained by 50 students in a test. We can find the cumulative frequency of each class interval and write it as,

Marks	Number of Students	Cumulative Frequency
0-10	8	8
10-20	4	12
20-30	10	22
30-40	12	34
40-50	13	47
50-60	3	50

It is given that $60\%$ of the students pass the test. So, we can say that $40\%$ of the students has failed the test. So, in the ascending order, we can say that the mark after the $40\%$ of the entry is the minimum marks obtained by the pass student.
Now we can find the class at which the required term lies.
$\Rightarrow N = 40\% \times n$
It is given that the marks are of 50 students.
$\Rightarrow N = \dfrac{{40}}{{100}} \times 50$
On simplification, we get,
$\Rightarrow N = 20$
From the cumulative frequency, the required class is, 20-30.
Now the required mark is given by,
${M_{\min }} = l + \left( {\dfrac{{N - c.f.}}{f}} \right) \times h$
Here, l is the lower limit of the selected class which is 20,
c.f., is the cumulative frequency of the class preceding the selected class, which is 12,
f is the frequency of the selected class which is 10 and,
h is the class size which is 10.
On substituting these on the equation, we get,
$\Rightarrow {M_{\min }} = 20 + \left( {\dfrac{{20 - 12}}{{10}}} \right) \times 10$
On simplification, we get,
$\Rightarrow {M_{\min }} = 20 + 8$
Hence, we have,
$\Rightarrow {M_{\min }} = 28$
Therefore, the minimum marks obtained by a pass student is 28.

Note:
The grouped data is given in the ascending order. So, we can find the lower limit by simply finding the cumulative frequency and the number of the students failed. If we take the number of passed students, we must arrange the data in decreasing order, or we need to find the greater than cumulative frequency. The median of a grouped data is given by,
$Median = l + \left( {\dfrac{{\dfrac{n}{2} - c.f.}}{f}} \right) \times h$
Here, l is the lower limit of the median class,
n is the number of observations,
c.f., is the cumulative frequency of the class preceding the median class,
f is the frequency of the median class and,
h is the class size.