Answer: Option B. -> Non-response error
:
B
Non-response erroroccurs when sampling units selected for a sample are not interviewed. The non-response may be due to unwillingness or unavailability. It is a non-sampling error.

Answer: Option B. -> ¯x+n+1
:
B
$\overline{x}=\frac{{\sum }_{i=1}^n{x}_i}{n}\Rightarrow {\sum }_{i=1}^n{x}_i=n\overline{x}\therefore \frac{{\sum }_i^n(x_i+2i)}{n}=\frac{{\sum }_{i=1}^n{x}_i+2(1+2+...+n)}{n}=\frac{n\overline{x}+2\frac{n(n+1)}{2}}{n}=\overline{x}+(n+1)$

Answer: Option C. -> 6
:
C
$6=\frac{3+4+x+7+10}{5}$
$\Rightarrow 30=24+x$
$\Rightarrow x=6$

Answer: Option D. -> 1
:
D
We know that if we multiply all numbers of a set by a constant "k" then the mean of these numbers will also be a multiple of the old mean.
And the variance of these numbers will be multiple of "$k^2$".
Since standard deviation is nothing but the square root of variance, s.d will be multiple of "k" of the old standard deviation.
We are given the s.d was 4.
So after multiplying each observation by (1/4), the s.d will also be multiplied by (1/4), thus giving the value = 4$\times \frac{1}{4}$ = 1

Answer: Option C. -> Team A is more consistent than team B.
:
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.

Answer: Option C. -> ¯¯¯x+n+12
:
C
Let, $x_1,x_2.......x_n$ be n items.
Then, $\overline{x}=\frac{1}{n}\sum x_i$
$Let{y}_1=x_1+1,y_2=x_2+2,y_3=x_3+3,.....,y_n=x_n+n$
Then the mean of the new series is
$\frac{1}{n}\sum y_i=\frac{1}{n}\sum _{i=1}^n(x_i+i)$
$=\frac{1}{n}\sum _{i=1}^n{x}_i+\frac{1}{n}(1+2+3+......+n)$
$=\overline{x}+\frac{1}{n}.\frac{n(n+1)}{2}=\overline{x}+\frac{n+1}{2}$

Answer: Option D. -> remains the same as that of the original set
:
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 $\frac{n+1}{2}th$ 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.

Answer: Option C. -> all values of data are equal
:
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)
$\therefore$ M = m
$\therefore$ m $⩽$ each data item $⩽$ M = m
$\Rightarrow$ each data item = m or M i.e, all values of the data are equal.

Answer: Option A. -> 100×s.dmean
:
A
$C.V.isdefinedbyC.V=100\times \frac{s.d}{mean}$

Answer: Option C. -> 4 and 7
:
C
$\overline{x}=4,N=5\text{and}\frac{\sum (x-{\overline{x}}^2)}{N}=5.2\Rightarrow \sum (x-\overline{x}{)}^2=(5.2)5\therefore \sum (x-\overline{x}{)}^2=26\therefore (1-4{)}^2+(2-4{)}^2+(6-4{)}^2+(\alpha -4{)}^2+(\beta -4{)}^2=26\therefore (\alpha -4{)}^2+(\beta -4{)}^2=9\text{Also,}\frac{1+2+6+\alpha +\beta }{5}=4\therefore \alpha +\beta =20-9=11\text{Clearly 4, 7 only satisfy the above equation in}\alpha ,\beta .\text{Hence required numbers are 4, 7.}$