Question

In a continuous frequency distribution, class mark of a class is 85 and the class size is 6. Find
the corresponding class interval.

Answer 1

SOLUTION

GIVEN

In a continuous frequency distribution, class mark of a class is 85 and the class size is 6.

TO DETERMINE

The corresponding class interval.

CONCEPT TO BE IMPLEMENTED

CLASS INTERVAL

When the number of sample values are large in number and are in wide range, then the whole data is divided in a number of several groups according to the size of the sample. Each of these groups, made as above, is known as class interval

CLASS BOUNDARY

When class intervals are such a type that the upper class limit of any class is not equal to lower class limit of successive class then to get any continuous graphical representation of the data it is sometimes required to rearrange the class limits in such a way that the upper class limit of any class coincides with the lower class limit of next class. Then these class limits are called class boundaries. The lower and upper ends of any class are called lower and upper class boundaries respectively.

CLASS LENGTH

Length of the class interval of an class is defined to be the difference between the lower and upper class boundaries ( not class limits) of that particular class interval.

Class length = Upper Class boundary - Lower class boundary

CLASS FREQUENCY

The number of observed values falling within a class is called the frequency of that class

CLASS MARK

For any grouped frequency table with class intervals, the middle value of the class limits or the class boundaries of any class is called class mark of the class

EVALUATION

Here it is given that in a continuous frequency distribution, class mark of a class is 85 and the class size is 6

∴ Upper Class boundary

$= \displaystyle \sf{85 - \frac{6}{2} }$

= 85 - 3

= 82

∴ Lower class boundary

$= \displaystyle \sf{85 + \frac{6}{2} }$

= 85 + 3

= 88

Hence the required corresponding class interval is 82 - 88

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