Answer: Option A. -> True
:
A
The formula for the mode is as follows.
$Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$
$l=$ lower boundary of the modal class
$h=$ size of the modal class interval
$f_1=$ frequency of the modal class.
$f_0=$ frequency of the class preceding the modal class
$f_2=$frequency of the class succeeding the modal class
If the preceding and succeeding classes have the same frequency, then $f_0=f_2=f\left(say\right)$.
Then the equation reduces to
$Mode=l+\left(\frac{f_1-f}{2f_1-f-f}\right)\times h$
$Mode=l+\left(\frac{f_1-f}{2(f_1-f)}\right)\times h$
$Mode=l+\frac{1}{2}\times h$, which is the midpointof the modal class.
$\therefore$If the preceding and succeeding classes of the modal class have the same frequency, then the mode will be at the midpoint of the modal class.

Answer: Option B. -> False
:
B
The class mark is the midpoint of the lower and upper limits of a class. It is calculated by taking the average of the upper and lower limits i.e.
Class mark = $\frac{Upperlimit+Lowerlimit}{2}$

Answer: Option A. -> Most frequent value
:
A
Mode is the most frequently observed value

Answer: Option A. -> 4
:
A
We know that,
3 Median = Mode + 2 Mean
or
Mode = 3 Median - 2 Mean
= $3\times 6-2\times 7$
= 18 - 14
= 4
Therefore, mode = 4

Answer: Option D. -> 9
:
D
$\begin{array}{ccc}\text{Class}& \text{Frequency}& CumulativeFrequency\\ \text{0 - 20}& \text{10}& \text{10}\\ \text{20 - 40}& \text{7}& \text{17}\\ \text{40 - 60}& \text{16}& \text{33}\\ \text{60 - 80}& \text{p}& \text{33+p}\\ \text{80 -100}& \text{8}& \text{41+p}\end{array}$
Since median is 50, median class is 40-60.
Using
$Median=l+\left[\frac{\frac{n}{2}-cf}{f}\right]\times h$,
$Here,n=41+p,$
$⟹\frac{n}{2}=\frac{(41+p)}{2}$
$Now,50=40+[\frac{\frac{(41+p)}{2}-17}{16}\times 20]$
$⟹10=[\frac{\frac{(41+p-34)}{2}}{16}\times 20]$
$⟹10=[\frac{(7+p)}{16\times 2}\times 20]$
Simplifying, we get $\frac{(7+p)}{16}=1$
$7+p=16$
$p=9$

Answer: Option A. -> xi−A
:
A
In statistics, 'theassumed mean' is a method for calculating the arithmeticmeanand standard deviation of a data set. It simplifies calculating accurate values by hand.
During the application of the short-cut method for finding the mean, the deviation d, are divisible by a common number ‘h’.In this case, the di = xi – A is reduced to a great extent as 'di' becomes $\frac{di}{h}$.In this method,we divide the deviations by the same number to simplify calculations.The deviation is $d_i=x_i-A$

Answer: Option B. -> 20 - 30
:
B
The given table can be written as:
$$\begin{array}{cccccc}\text{Marks}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \text{No. of students}& 5& 5& 13& 8& 9\\ \text{CumulativeFrequency}& 5& 10& 23& 31& 40\end{array}$$
No. of students can be calculated as follows:
5
10 - 5 = 5
23 - 10 = 13
31 - 23 = 8
40 - 31 = 9
$\therefore \frac{n}{2}=\frac{40}{2}=20$
Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.
Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.
The class interval 20 - 30 is the required answer.

Answer: Option C. -> 50
:
C
Mean = 45 and Mode = 60
3 Median = Mode +2 Mean
3 Median = 60 +2(45)
3 Median = 150
Median=50

Answer: Option A. -> l+(f1−f02f1−f0−f2) ×h
:
A
$l+(\frac{f_1-f_0}{2f_1-f_0-f_2}$) $\times h$is used to determine the mode .
A mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency.

:
$$\begin{array}{ccc}{x}_i& f_i& f_i{x}_i\\ \text{10}& \text{10}& \text{100}\\ \text{p}& \text{15}& \text{15p}\\ \text{18}& \text{7}& \text{126}\\ \text{21}& \text{9}& \text{189}\\ \text{25}& \text{9}& \text{225}\end{array}$$
$\sum f_i$=50
$\sum f_i{x}_i$ = 640+15p
Given mean=17
17 = $\frac{640+15p}{50}$
Solving, we get p = 14