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The class marks of a distribution are 61,66,71,76,81,86,91,96,101,106. Determine the class size, class limits and true class limits.

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Since class marks (mid values) are uniformly spaced, so the class-size is the differnce between any two consecutive class marks. Therefore, class size = 66-61 = 5
Let the lower limit of the first class be a.
`therefore "Its upper limit" =a+5`
Its mid value = 61
`therefore " Class mark" = "upper limit + lower limit"/(2)`
`rArr (a+(a+5))/(2)=61`
`implies a+(a+5)=122`
`implies 2a+5 = 122 rArr 2a-117 rArr a=58.5`
`therefore" The first class interval is " 58.5-63.5. " So, class intervals are "`
image
The above is the exclusive series. So, the true class limits are same as class limits.

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