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

$$\begin{array}{ccc}X& F& FX\\ \text{10}& \text{10}& \text{100}\\ \text{p}& \text{15}& \text{15p}\\ \text{18}& \text{7}& \text{126}\\ \text{21}& \text{9}& \text{189}\\ \text{25}& \text{9}& \text{225}\end{array}$$
$\sum F=50$
$\sum FX=640+15p$
Given mean=17
$17=\frac{640+15p}{50}\Rightarrow 15p=50\times 17-640=210\Rightarrow p=\frac{210}{15}=14$

Given that the modal class is 20- 40
Frequency of modal class= 8
Frequency of the preceding class = 7
Frequency of the succeeding class = 2
$l=20;D_1=8-7=1$
$D_2=8-2=6;h=20$
$Mode=l+\left(\frac{D_1}{D_1+D_2}\right)\times h=20+\frac{1}{7}\times 20=22.86$

The median for grouped data is formed by $l+\left(\frac{\frac{N}{2}-CF}{f}\right)\times h$.

Where l is the lower class limit of the median class, N is the total number of observations, CF is the cumulative frequency of the class preceding the median class, f is the frequency of the median class and h is the class size.

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.

The first 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29

The sum of numbers = 129

The Mean = 129/10 = 12.9

$\begin{array}{ccc}\text{Class Interval}& \text{Frequency}& \text{Cumulative Frequency}\\ 100-120& 19& 19\\ 120-140& 28& 47\\ 140-160& 30& 77\\ 160-180& 16& 93\\ 180-200& 7& 100\end{array}$
$\frac{N}{2}=50$
Median class is 140-160
$M=L+\frac{\frac{N}{2}-c.f}{f}\times h=140+\frac{50-47}{30}\times 20=142$

The number of athletes who completed the race in less than 14.6 is given by the cumulative frequency of the class 14.4 - 14.6

= 2 + 4 + 5 + 71

= 82