
Using an appropriate method, find the mean of following frequency distribution:
Class 84 - 90 90 - 96 96 - 102 102-108 108 - 114 114 - 120 Frequency 8 10 16 23 12 11
Which method did you use, and why?
Class | 84 - 90 | 90 - 96 | 96 - 102 | 102-108 | 108 - 114 | 114 - 120 |
Frequency | 8 | 10 | 16 | 23 | 12 | 11 |
Answer
487.2k+ views
Hint: The mean (or average) of observations, as we know is the sum of the values of all the observations divided by the total number of observations. If are observation with respective frequencies , then their mean observation occurs time, occurs time and so on.
So, the mean of the data is give by
Recall that we can write this in short form by using the Greek letter which means summation.
Varies from to .
Complete step-by-step answer:
It is assumed that the frequency of each class – Internal is centered around its main point. So the midpoint (class marks) of each class can be chosen to represent the observation in the case.
Class mark =
We have the sigma of frequencies is and sigma is is . So, the mean of the given data is the give by
The mean value is
The choice of method to be used depends on the numerical value of and .
If and are sufficiently small, then the direct method is an appropriate choice.
Note: If and are numerically large numbers, then we can go for the assumed mean method of step – deviation method. If the class size is unequal and are large. Numerically, we can still apply the step-deviation method for te mean by taking h to be a suitable divisor of all the
So, the mean
Recall that we can write this in short form by using the Greek letter
Complete step-by-step answer:
class Interval | Frequency( | Class mark( | |
It is assumed that the frequency of each class – Internal is centered around its main point. So the midpoint (class marks) of each class can be chosen to represent the observation in the case.
Class mark =
We have the sigma of frequencies is
The mean value is
The choice of method to be used depends on the numerical value of
If
Note: If
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