Question

Find the actual lower and upper class limits and also the class marks of the classes :
$1.1-2.0,2.1-3.0$ and $3.1-\mathrm{4.0.}$

• A. Actual lower limits : $1.05,2.05,3.05$
  Actual upper limits : $2.05,3.05,4.05$
  Class marks : $1.55,2.55,3.55$
• B. Actual lower limits : $1.5,2.5,3.5$
  Actual upper limits : $2.5,3.5,4.5$
  Class marks : $2.5,3.5,4.5$
• C. Data insufficient
• D. None of these

Answer 1

Answer: (A)

Actual lower limits : $1.05,2.05,3.05$

Actual upper limits : $2.05,3.05,4.05$
Class marks : $1.55,2.55,3.55$

$d=\frac{LowerLimit-upperLimitOfPreviousColumn}{2}=\frac{2.1-2.0}{2}=0.05$

Class mark is the mid point of the class interval.

Upper limit is highest value of the class interval and the actual upper limit is obtained by adding $0.5$ to the highest number if number is represented as whole number or add $0.05$ to the highest number if the number is represented as decimal.

Similarly, the lower limit is the smallest value of the class interval and the actual lower limit is obtained by subtracting $0.5$ to the smallest number if the number is whole number or subtract $0.05$ to the smallest number if the number is decimal.

Given, class intervals are $1.1-2.0,2.1-3.0,3.1-4.0$

Since, the intervals are represented in decimal, we add and subtract $0.05$ to get the actual limits.

For class interval $1.1-2.0$,

Class mark is $\frac{1.1+2.0}{2}=\frac{3.1}{2}=1.55$

Actual upper limit is $2.0+0.05=2.05$ and

actual lower limit is $1.1-0.05=1.05$

Similarly for $2.1-3.0$,

Class mark is $\frac{2.1+3.0}{2}=2.55$

Actual upper limit is $3.0+0.05=3.05$ and

actual lower limit is $2.1-0.05=2.05$

Similarly for $3.1-4.0$,

Class mark is $\frac{3.1+4.0}{2}=3.55$

Actual upper limit is $4.0+0.05=4.05$ and

actual lower limit is $3.1-0.05=3.05$

Therefore, actual upper limits are $2.05,3.05,4.05$

Actual lower limits are $1.05,2.05,3.05$

Class marks are $1.55,2.55,3.55$