What is the class mark of the class interval –(80 – 90)?
(a) 82.5
(b) 90
(c) 80
(d) 85
Answer
601.8k+ views
Hint: To solve this question, we will first of all define what a class mark is then we would try to compute the process by which it is obtained. If the class range is between 0 – 10, then the classmark is given by \[\dfrac{0+10}{2}=\dfrac{10}{2}=5.\] And hence, we can exactly determine the class mark.
Complete step by step answer:
Let us first define class – mark is. Use class midpoint (or classmark) is a specific point in the center of the bins (categories) in a frequency distribution table. It is also the center of a bar in a histogram. It is defined as the average of the upper and the lower class limits.
Let us consider an example of Class Mark:
Let us look at how we got the class mark for each class interval.
\[\Rightarrow \dfrac{7+9}{2}=\dfrac{16}{2}=8\]
\[\Rightarrow \dfrac{10+12}{2}=\dfrac{22}{2}=11\]
\[\Rightarrow \dfrac{13+15}{2}=\dfrac{28}{2}=14\]
\[\Rightarrow \dfrac{16+18}{2}=\dfrac{34}{2}=17\]
\[\Rightarrow \dfrac{19+21}{2}=\dfrac{40}{2}=20\]
We can clearly understand that to calculate the mean of a continuous series, the midpoints of the various class intervals are taken before calculating the mean which is also called the class mark.
Now, coming to our question, we have been given the class interval as (80 – 90). So, we can find the class mark as \[\dfrac{80+90}{2}=\dfrac{170}{2}=85.\]
Therefore, the class mark of the class interval (80 – 90) is 85.
So, the correct answer is “Option D”.
Note: When class interval differs by a 10 difference means if the class interval is of the term
Then the classmark corresponding to it can be easily written as
If the class interval is not continuous, then we first make it continuous and then calculate the class mark by the average method.
Complete step by step answer:
Let us first define class – mark is. Use class midpoint (or classmark) is a specific point in the center of the bins (categories) in a frequency distribution table. It is also the center of a bar in a histogram. It is defined as the average of the upper and the lower class limits.
Let us consider an example of Class Mark:
| Class Interval | Class Mark |
| 7 – 9 | 8 |
| 10 – 12 | 11 |
| 13 – 15 | 14 |
| 16 – 18 | 17 |
| 19 – 21 | 20 |
Let us look at how we got the class mark for each class interval.
\[\Rightarrow \dfrac{7+9}{2}=\dfrac{16}{2}=8\]
\[\Rightarrow \dfrac{10+12}{2}=\dfrac{22}{2}=11\]
\[\Rightarrow \dfrac{13+15}{2}=\dfrac{28}{2}=14\]
\[\Rightarrow \dfrac{16+18}{2}=\dfrac{34}{2}=17\]
\[\Rightarrow \dfrac{19+21}{2}=\dfrac{40}{2}=20\]
We can clearly understand that to calculate the mean of a continuous series, the midpoints of the various class intervals are taken before calculating the mean which is also called the class mark.
Now, coming to our question, we have been given the class interval as (80 – 90). So, we can find the class mark as \[\dfrac{80+90}{2}=\dfrac{170}{2}=85.\]
Therefore, the class mark of the class interval (80 – 90) is 85.
So, the correct answer is “Option D”.
Note: When class interval differs by a 10 difference means if the class interval is of the term
| Class Interval |
| 0-10 |
| 10-20 |
| 20-30 |
| 30-40 |
| 40-50 |
Then the classmark corresponding to it can be easily written as
| Class Interval | Class Mark |
| 0 – 10 | 5 |
| 10 – 20 | 15 |
| 20 – 30 | 25 |
| 30 – 40 | 35 |
| 40 – 50 | 45 |
If the class interval is not continuous, then we first make it continuous and then calculate the class mark by the average method.
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