Here, we see that the difference between two mid-points is, 15 - 5 i.e., 10.
It means the width of the class interval is 10.
Let the lower limit of the first class interval be a.
Then, its upper limit = a + 10
Now, mid value of the first class interval = 5
⇒ mid value =Lower limit + Upper limit2
⇒5=a+a+102⇒2a+10=10
⇒2a=0⇒a=0
So, the first class interval is 0 - 10.
Now, we prepare a continuous grouped frequency distribution table is given below.
Mid-pointClass intervalFrequency50−1041510−2082520−30133530−40124540−506
The size of the class interval is 10.