# Free Objective Test 02 Practice Test - 11th and 12th

If the range of discrete data of n observations is zero, then

A.

all values of data are zero

B.

all values of data are equal to standard deviation

C.

all values of data are equal

D.

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.

If the median and mode of 4 observations is 4, 6 and the sum of squares of observations is 48.  Then
standard deviation is

A. 2
B. 4
C. 3
D. 5

#### SOLUTION

Solution : C

Given  that  median = 4  and  mode = 6.  We  know  that
Mean - mode = 3  (Mean - Median)
Mean - 6 = 3 (Mean - 4)
¯x6=3¯x12
¯x=3
σ2=x2in(xin)2=484(3)2=3 σ=3

The coefficient of  variation of  a distribution is

A. 100×s.dmean
B. 100×means.d
C. 100×variancemean
D. s.dmean

#### SOLUTION

Solution : A

C.V. is defined byC.V=100×s.dmean

The mean deviation of 2, 4, 6, 8, 10 about its mean is

A. 4.2
B. 2.4
C. 3.4
D. 4.3

#### SOLUTION

Solution : B

Mean(¯x)=xin=2+4+6+8+105=6 M.D=|xi¯x|n=|26|+|46|+|66|+|86|+|106|5=4+2+0+2+45=125=2.4

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

A. is increased by 2
B. is decreased by 2
C. is two times the original median
D. remains the same as that of the original 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.

If a variable takes the discrete values α+4,α72,α52,α3,α2,α+12,α12,α+5(α>0),then the median is

A. α54
B. α12
C. α2
D. α+54

#### SOLUTION

Solution : A

Arrange the  data  as:  α72,α3,α52,α2,α12,α+12,α+4,α+5,

Median =a2+α122=2α522=α54

The variance of the data 2, 4, 6, 8, 10 is:

A. 6
B. 7
C. 8
D. none of these

#### SOLUTION

Solution : C

Mean(¯x)=xin=((2+4+6+8+10)5=6

Variance(σ2)=(x¯x)2n=(26)2+(46)2+(66)2+(86)2+(106)25=(16+4+0+4+16)5=405=8

The mean of a set of observation is ¯¯¯x .If each observation is divided by α α0, and then is increased by 10,
then the mean of the new set is

A. ¯¯¯xα
B. ¯¯¯x+10α
C. ¯¯¯x+10αα
D. α¯¯¯x+10

#### SOLUTION

Solution : C

Let x1,x2,xn be n observations.
Then ¯¯¯x=1nxi
let yi=xiα+10
¯¯¯y=1nniyi
And, ¯¯¯y=1n[(nixiα+10.n]
¯¯¯y=1α¯¯¯x+10
=¯¯¯x+10αα.

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

A. 75
B. 157
C. 82
D. 80

#### SOLUTION

Solution : A

Given,
82=(27+x)+(31+x)+(89+x)+(107+x)+(156+x)5
82×5=410+5x410410=5xx=0
Required mean is,
¯¯¯x=130+x+126+x+68+x+50+x+1+x5
¯¯¯x=375+5x5=375+05=3755=75.

If the algebraic sum of deviations of 20 observations from 30 is 20, then the mean of observations is

A. 30
B. 30.1
C. 29
D. 31

#### SOLUTION

Solution : D

20i=1(xi30)=20=20i=1xi20×30=20
20i=1xi=620. Mean =20i=120=62020=31.

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

A. 55 kg
B. 60 kg
C. 57 kg
D. 50 kg

#### 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 =385328=57kg.

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

A.

None of the above

B.

¯¯¯x+n2

C.

¯¯¯x+n+12

D.

¯¯¯x+n

#### SOLUTION

Solution : C

Let, x1,x2.......xn be n items.
Then, ¯¯¯x=1nxi
Let y1=x1+1,y2=x2+2,y3=x3+3,.....,yn=xn+n
Then the mean of the new series is
1nyi=1nni=1(xi+i)
=1nni=1xi+1n(1+2+3+......+n)
=¯¯¯x+1n.n(n+1)2=¯¯¯x+n+12

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

A. 14.98, 39.95
B. 39.95, 14.98
C. 39.95, 224.5
D. None the these

#### SOLUTION

Solution : B

Corrected x=40×20050+40=7990
Corrected ¯¯¯x=7990200=39.95
x2=n[σ2+¯¯¯x2]=200[152+402]=365000
Correct x2=3650002500+1600=364100
Corrected σ=364100200(39.95)2
=(1820.51596)=224.5=14.98.

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?

A. Team B is more consistent than team A.
B. Two teams are equally consistent.
C. Team A is more consistent than team B.
D.  The s.ds. of team A and team B are 125,172.

#### 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.

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

A.

49

B.

23

C.

1

D.

94

#### SOLUTION

Solution : C

Population B observations are 151, 152, ….., 250
Let y, be these observations
Shifting the origin zi=yi50
The values of these observations are 100, 101, ….., 200.
VA=VB
( Variance is independent of shifting the origin).