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1.

The range of the data 12,25,15,18,17,20,22,6,16,11,8,19,10,30,20,32, isA. 10B. 15C. 18D. 26

Answer» Correct Answer - D
Range=(maximum value)-(minimum value)=(32-6)=26
Discussion
2.

The class marks of a frequency distribution are given as follows 15,20, 25,……….. The class corresponding to the class mark 20 isA. 12.5-17.5B. 17.5-22.5C. 18.5-21.5D. 19.5-20.5

Answer» Correct Answer - B
Class size=(20-15)=5
Class mark=20.
Lower limit=`(20-(5)/(2))=(35)/(2)=17.5`
Upper limit=`(20+(5)/(2))=(45)/(2)=22.5`
Required class is 17.5-22.5
Discussion
3.

If m is the mid-point and l is the upper limit of a class in a continous frequency distribution, then lower class limit of the class isA. 2m-uB. 2m+uC. m-uD. m+u

Answer» Correct Answer - A
Let l be the lower limit. Then,
`m=(l+u)/(2)Rightarrow l=2m-u`
Discussion
4.

(i)Find the class mark of the class 90-120. (ii)In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lowe limit of the class. (iii)The width of each of the five continuous classes in a frequency distribution is 5 and lowe class limit of the lowest class is 10. What is the upper class limit of the highest class? (iv) The class marks of a frequency distribution are 15,20,25...... Find the class corresponding to the class mark 20. (v)In the class intervals 10-20, 20-30, find the class in which 20 is included.

Answer» (i)105 (ii)7 (iii)35 (iv)17.5-22.5 (v)20-30
Discussion
5.

In a frequency distribution, the mid values of a class is 10 and width of the class is 6. The lower limit of the class isA. 6B. 7C. 8D. 12

Answer» Correct Answer - B
Let the lower limit be a. Then, upper limit=a+3.
Mid value of `=(a+a+6)/(2)=(2a+6)/(2)=a+3`
`therefore a+3=10 Rightarrow a=7`
`therefore` lower limit of the class is 7.
Discussion
6.

The mid value of a class interval in 42 and the class size is 10 then find lower and upper limits.A. 37-47B. 37.5-47.5C. 36.5-47.5D. 36.5-46.5

Answer» Correct Answer - A
Let the lower limit be x. Then upper limit=(x+10).
`therefore (x(x+10))/(2)=42 Rightarrow 2x+10=84 Rightarrow 2x=84 Rightarrow x=37`
`therefore` lower limit=37 and upper limit=47
Discussion