Let x and y be the upper and lower class limit of frequency distribution.
Given, width of the class = 5
⇒ x-y= 5 …
Also, given lower class (y) = 10 On putting y = 10, we get
x – 10= 5 ⇒ x = 15 So, the upper class limit of the lowest class is 15
Hence, the upper class limit of the highest class
=(Number of continuous classes x Class width + Lower class limit of the lowest class)
= 5 x 5+10 = 25+10=35
Hence,’the upper class limit of the highest class is 35.
<em><u>Alternate Method</u></em><em><u>.</u></em>
After finding the upper class limit of the lowest class, the five continuous classes in a frequency distribution with width 5 are 10-15,15-20, 20-25, 25-30 and 30-35.
Thus, the highest class is 30-35..
<h3>Hence, the upper limit of this class is 35.</h3>
<h2>_____________________</h2>
, A value of 5 would make the denominator 0. This is undefined, so 5 is excluded.