Free Statistics 01 Practice Test - 10th Grade 

Class mark is the midpoint of a class interval.
Therefore, its formula is given by $\frac{\text{upper limit + lower limit}}{2}$.

$$\begin{array}{ccc}{x}_i& f_i& f_i{x}_i\\ \text{10}& \text{2}& \text{20}\\ \text{13}& \text{5}& \text{65}\\ \text{16}& \text{7}& \text{112}\\ \text{19}& \text{6}& \text{114}\\ & \sum f_i=20& \sum f_i{x}_i=311\end{array}$$

Mean $=\frac{\sum f_i{x}_i}{\sum f_i}$
         
          $=\frac{311}{20}$ 
         
          $=15.55$

$$\begin{array}{ccc}{x}_i& f_i& f_i{x}_i\\ \text{10}& \text{10}& \text{100}\\ \text{p}& \text{15}& \text{15p}\\ \text{18}& \text{7}& \text{126}\\ \text{21}& \text{9}& \text{189}\\ \text{25}& \text{9}& \text{225}\end{array}$$

$\sum f_i$=50
$\sum f_i{x}_i$ = 640+15p
Given mean=17
17 = $\frac{640+15p}{50}$
Solving, we get p = 14

 $\begin{array}{ccc}\text{Class}& \text{Frequency}& CumulativeFrequency\\ \text{0 - 20}& \text{10}& \text{10}\\ \text{20 - 40}& \text{7}& \text{17}\\ \text{40 - 60}& \text{16}& \text{33}\\ \text{60 - 80}& \text{p}& \text{33+p}\\ \text{80 -100}& \text{8}& \text{41+p}\end{array}$

 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 $\frac{(7+p)}{16}=1$
$7+p=16$
$p=9$

Mode is the most frequently observed value

In the formula for mode, $f_1$ represents the frequency of the modal class.

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 = $\frac{Upperlimit+Lowerlimit}{2}$

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 midpoint of 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.

We know Mean = $\frac{\sum fx}{\sum f}$

Hence, $\frac{5x+2}{12}$ = 6 

$\Rightarrow$ 5x + 2 = 12 $\times$ 6
$\Rightarrow$ 5x = 72 – 2 = 70
$\Rightarrow$ x = 14