Such series may be converted into simple frequency series using the following method: (i) First, mutual difference between mid-values (i), is determined, and (ii) Second, the difference so obtained is reduced to half (1/2i) which when deducted from the mid-value gives lower limit of the class interval and when added to the mid-value gives the corresponding upper limit.

Such series may be converted into simple frequency series using the

1.	Such series may be converted into simple frequency series using the following method: (i) First, mutual difference between mid-values (i), is determined, and (ii) Second, the difference so obtained is reduced to half (1/2i) which when deducted from the mid-value gives lower limit of the class interval and when added to the mid-value gives the corresponding upper limit.
Answer» Solution :Thus,Lower limit: `l_1 = m-1/2i` Upper limit: `l_2=m+1/2i` where, m=mid-value, i=difference between mid-values, `l_1`=lower limit and `l_2` = upper limit. In the above frequency series with mid values, the MUTUAL difference between mid-values (i) 15 - 5=10. HALF of it is 5. Deducting 5 from each mid-value we GET lower limits and adding 5 to each mid-value we get the corresponding upper limits. Following table shows the process of this conversion.