
What is the class interval of the fourth class?
score frequency Below $75$ $7$ $76 - 80$ $17$ $81 - 85$ $7$ $86 - 90$ $6$ $91 - 95$ $6$ $96 - 100$ $1$
A. $6$
B. $5$
C. $4$
D. $1$
score | frequency |
Below $75$ | $7$ |
$76 - 80$ | $17$ |
$81 - 85$ | $7$ |
$86 - 90$ | $6$ |
$91 - 95$ | $6$ |
$96 - 100$ | $1$ |
Answer
515.4k+ views
Hint:
This is not the continuous distribution so first of all we need to convert it into the continuous distribution by adding and subtracting the value of $0.5$ from the upper and the lower limit. Then we’ll check the required fourth class distribution.
Complete step by step solution:
Here we are given the distribution of the score and the frequency.
So here we see that the score is given and the frequency corresponding to that mark is also given. So here we are told that if the marks are less than $75$ then the number of students scoring them is $7$ and similarly we are given other intervals also and the marks obtained by the number of students in them.
But the problem in solving it is that it is not the continuous distribution as in the continuous distribution we have the upper limit of the first interval is the same as the lower limit of the next succeeding class interval.
But here it is not the same. So we need to convert it into the continuous distribution by adding and subtracting the value of $0.5$ from the upper and the lower limit.
Here the frequency of $75 - 76,80 - 81,85 - 86,95 - 96$ is missing. So here first of all we need to take the average of the upper limit of one class interval and lower limit of the other class interval which is $\dfrac{{75 + 76}}{2} = 75.5$ and similarly we get the other terms as $80.5, 85.5, 90.5, 95.5, 100.5$
So we can write the distribution table as
So this becomes the continuous form and we are asked to find the class interval of the fourth class which is here $85.5 - 90.5$
So the class interval; of the 4th class interval is $90.5 - 85.5 = 5$
So option B is correct.
Note:
If we are asked to find the modal class then the class with the maximum frequency is termed as the modal class that is $75.5 - 80.5$ as it is the modal class with the frequency $17$. It is always required to have continuous distribution to check the model class frequency.
This is not the continuous distribution so first of all we need to convert it into the continuous distribution by adding and subtracting the value of $0.5$ from the upper and the lower limit. Then we’ll check the required fourth class distribution.
Complete step by step solution:
Here we are given the distribution of the score and the frequency.
score | frequency |
Below $75$ | $7$ |
$76 - 80$ | $17$ |
$81 - 85$ | $7$ |
$86-90$ | $6$ |
$91 - 95$ | $6$ |
$96 - 100$ | $1$ |
So here we see that the score is given and the frequency corresponding to that mark is also given. So here we are told that if the marks are less than $75$ then the number of students scoring them is $7$ and similarly we are given other intervals also and the marks obtained by the number of students in them.
But the problem in solving it is that it is not the continuous distribution as in the continuous distribution we have the upper limit of the first interval is the same as the lower limit of the next succeeding class interval.
But here it is not the same. So we need to convert it into the continuous distribution by adding and subtracting the value of $0.5$ from the upper and the lower limit.
Here the frequency of $75 - 76,80 - 81,85 - 86,95 - 96$ is missing. So here first of all we need to take the average of the upper limit of one class interval and lower limit of the other class interval which is $\dfrac{{75 + 76}}{2} = 75.5$ and similarly we get the other terms as $80.5, 85.5, 90.5, 95.5, 100.5$
So we can write the distribution table as
score | frequency |
Below $75.5$ | $7$ |
$75.5 - 80.5$ | $17$ |
$80.5 - 85.5$ | $7$ |
$85.5 - 90.5$ | $6$ |
$90.5 - 95.5$ | $6$ |
$95.5 - 100.5$ | $1$ |
So this becomes the continuous form and we are asked to find the class interval of the fourth class which is here $85.5 - 90.5$
So the class interval; of the 4th class interval is $90.5 - 85.5 = 5$
So option B is correct.
Note:
If we are asked to find the modal class then the class with the maximum frequency is termed as the modal class that is $75.5 - 80.5$ as it is the modal class with the frequency $17$. It is always required to have continuous distribution to check the model class frequency.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Trending doubts
1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Explain zero factorial class 11 maths CBSE
