Free Objective Test 02 Practice Test - 11th and 12th
Question 1
If the range of discrete data of n observations is zero, then
all values of data are zero
all values of data are equal to standard deviation
all values of data are equal
the extreme values of data are different
SOLUTION
Solution : C
Let m be the least value of the discrete data and M be the maximum value of the discrete data.
Then range = M – m = 0 (hypothesis)
∴ M = m
∴ m ⩽ each data item ⩽ M = m
⇒ each data item = m or M i.e, all values of the data are equal.
Question 2
If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48. Then
standard deviation is
SOLUTION
Solution : C
Given that median = 4 and mode = 6. We know that
Mean - mode = 3 (Mean - Median)
Mean - 6 = 3 (Mean - 4)
¯x−6=3¯x−12
¯x=3
σ2=∑x2in−(∑xin)2=484−(3)2=3 ∴σ=√3
Question 3
The coefficient of variation of a distribution is
SOLUTION
Solution : A
C.V. is defined byC.V=100×s.dmean
Question 4
The mean deviation of 2, 4, 6, 8, 10 about its mean is
SOLUTION
Solution : B
Mean(¯x)=∑xin=2+4+6+8+105=6∴ M.D=∑|xi−¯x|n=|2−6|+|4−6|+|6−6|+|8−6|+|10−6|5=4+2+0+2+45=125=2.4
Question 5
The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations in the set is increased by 2, then the median of the new set
SOLUTION
Solution : D
We know that median is found when we arrange the observations in ascending or descending order. And if the number of observations are odd then it is the n+12th term.
Since on arranging 9 observations 5th will be the median.
On changing 4 of any side won't affect the median and it'll remain same.
Question 6
If a variable takes the discrete values α+4,α−72,α−52,α−3,α−2,α+12,α−12,α+5(α>0),then the median is
SOLUTION
Solution : A
Arrange the data as: α−72,α−3,α−52,α−2,α−12,α+12,α+4,α+5,
Median =a−2+α−122=2α−522=α−54
Question 7
The variance of the data 2, 4, 6, 8, 10 is:
SOLUTION
Solution : C
Mean(¯x)=∑xin=((2+4+6+8+10)5=6
Variance(σ2)=∑(x−¯x)2n=(2−6)2+(4−6)2+(6−6)2+(8−6)2+(10−6)25=(16+4+0+4+16)5=405=8
Question 8
The mean of a set of observation is ¯¯¯x .If each observation is divided by α α≠0, and then is increased by 10,
then the mean of the new set is
SOLUTION
Solution : C
Let x1,x2……,xn be n observations.
Then ¯¯¯x=1n∑xi
let yi=xiα+10
¯¯¯y=1nn∑iyi
And, ¯¯¯y=1n[(n∑ixiα+10.n]
⇒¯¯¯y=1α¯¯¯x+10
=¯¯¯x+10αα.
Question 9
If the mean of the number 27+x,31+x,89+x,107+x,156+x is 82, then the mean of 130+x,126+x,68+x,50+x,1+x is
SOLUTION
Solution : A
Given,
82=(27+x)+(31+x)+(89+x)+(107+x)+(156+x)5
⇒82×5=410+5x⇒410−410=5x⇒x=0
∴ Required mean is,
¯¯¯x=130+x+126+x+68+x+50+x+1+x5
¯¯¯x=375+5x5=375+05=3755=75.
Question 10
If the algebraic sum of deviations of 20 observations from 30 is 20, then the mean of observations is
SOLUTION
Solution : D
20∑i=1(xi−30)=20=20∑i=1xi−20×30=20
⇒20∑i=1xi=620. Mean =20∑i=120=62020=31.
Question 11
The mean weight per student in a group of seven students is 55 kg if the individual weights of 6 students are 52, 58, 55, 53, 56 and 54; then weight of the seventh student is
SOLUTION
Solution : C
Total weight of 7 students is =55×7=385kg
Sum of weight of 6 students
=52+58+55+53+56+54=328kg
∴ Weight of seventh student =385–328=57kg.
Question 12
The mean of n items is ¯¯¯x. If the first term is increased by 1 second by 2 and so on, then new mean is
None of the above
¯¯¯x+n2
¯¯¯x+n+12
¯¯¯x+n
SOLUTION
Solution : C
Let, x1,x2.......xn be n items.
Then, ¯¯¯x=1n∑xi
Let y1=x1+1,y2=x2+2,y3=x3+3,.....,yn=xn+n
Then the mean of the new series is
1n∑yi=1nn∑i=1(xi+i)
=1nn∑i=1xi+1n(1+2+3+......+n)
=¯¯¯x+1n.n(n+1)2=¯¯¯x+n+12
Question 13
The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are
SOLUTION
Solution : B
Corrected ∑x=40×200−50+40=7990
∴Corrected ¯¯¯x=7990200=39.95
∑x2=n[σ2+¯¯¯x2]=200[152+402]=365000
Correct ∑x2=365000−2500+1600=364100
∴Corrected σ=√364100200−(39.95)2
=√(1820.5−1596)=√224.5=14.98.
Question 14
If the coefficient of variation of data pertaining to goals scored by two teams A and B in a football season are 2.5 and 7.2 respectively, then which of the following statements i strue?
SOLUTION
Solution : C
We know that, lesser is the coefficient of variation, the data is more consistent.
Since C.V. of the team A is 2.5, which is less than the C.V. of team B, we can conclude that team A is more Consistent.
Question 15
Suppose a population A has 100 observations 101, 102,……200 and another population B has 100 observation 151, 152….., 250. If VA,VB represent the variances of the two populations respectively, then VAVB is
49
23
1
94
SOLUTION
Solution : C
Population B observations are 151, 152, ….., 250
Let y, be these observations
Shifting the origin zi=yi−50
The values of these observations are 100, 101, ….., 200.
∴VA=VB
( ∵ Variance is independent of shifting the origin).