Question

in a continous frequency
distribution table if the mid point of a class is 42 and class size is 10 then let us write the upper and lower limit of the class.


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Answer 1

Answer:

Step-by-step explanation: Step-by-step explanation:

Let the upper and lower limits are x and y respectively.

Now, (x + y)/2 = 42

=> x + y = 42*2

=> x + y = 84

Again given,

class size = 10

=> x - y = 10 .

Add equation 1 and 2, we get

2x = 94

=> x = 94/2

=> x = 47

From equation 2

47 - y = 10

=> y = 47 - 10

=> y = 37

So, lower limit = 37

and upper limit = 47