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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» <html><body><p></p>Solution :Thus,Lower limit: `l_1 = m-1/2i` <br/>Upper limit: `l_2=m+1/2i` <br/>where, m=mid-value, i=difference between mid-values, `l_1`=lower limit and `l_2` = upper limit. <br/> In the above frequency series with mid values, the <a href="https://interviewquestions.tuteehub.com/tag/mutual-2850190" style="font-weight:bold;" target="_blank" title="Click to know more about MUTUAL">MUTUAL</a> difference between mid-values (i) 15 - <a href="https://interviewquestions.tuteehub.com/tag/5-319454" style="font-weight:bold;" target="_blank" title="Click to know more about 5">5</a>=10. <a href="https://interviewquestions.tuteehub.com/tag/half-1014510" style="font-weight:bold;" target="_blank" title="Click to know more about HALF">HALF</a> of it is 5. Deducting 5 from each mid-value we <a href="https://interviewquestions.tuteehub.com/tag/get-11812" style="font-weight:bold;" target="_blank" title="Click to know more about GET">GET</a> lower limits and adding 5 to each mid-value we get the corresponding upper limits. <br/> Following table shows the process of this conversion. <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/TRJ_ECO_XI_C04_S01_007_S01.png" width="80%"/></body></html>


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