Size of the track suit 40 has the maximum frequency with 37 persons.
\(\therefore\) The mode of the distribution = 40 

The five integers are 3, 4, 6, 9, x
Mean \(\bar{x}=\cfrac\Sigma{x}=\cfrac{3+4+6+9+x}{5}\)
=\(\cfrac{22+x}{5}\)
Their Median = 6
Mean = twice of the median
\(\cfrac{22+x}{5}=2\times6\)
22+x = 5 \(\times\) 12 = 60
\(\therefore\) x = 60- 22 = 38

Calculate the median

  N = 50
Median Class = \(\left(\cfrac{N}{2}\right)^{th}\)Value
=\(\left(\cfrac{50}{2}\right)^{th}\) Value = 25^th Value
Median value = 30 - 40
\(\cfrac{N}{2}\)=25, l = 30
m = 24, c =10, f =10
\(\therefore\) Median = \(\cfrac{l+\left(\cfrac{N}{2}-m\right)}{f}\times{c} \)
=\(\cfrac{30+25-24}{10}\times{10}=31\)

Step Deviation Method
\(\bar { x } =A+\left[ \frac { \Sigma fd }{ \Sigma f } \times c \right] \) Where \(d=\frac { x-A }{ c } \)
Let Assumed mean A = 32
Class Interval c = 45

Mean \(\bar { x } =A+\left[ \cfrac { \Sigma fd }{ \Sigma f } \times c \right] \)
\(=32+\left[ \cfrac { -62 }{ 105 } \times 5 \right] \)
= 32 + (-2.952)
= 29.05

Let Assumed mean be 25

Mean \(\bar{x}=A+\frac{\Sigma fd}{\Sigma f}\)
= \(25+\frac{270}{63}\)
= 25 + 4.29 = 29.29

Median class = \(\left(N\over 2\right)^{th}\) value=\(\left(200\over 2\right)^{th}\) value
= 100^th value
Median class = 2000 – 3000
\({N\over 2}=100\ l=2000\)
m = 74, c = 1000, f = 54
Median = \(l+{\left({n\over2}-m\right)\over f}\times c\)
\(=2000+\left(100-74\over54\right)\times1000\)
\(=2000+({26\over54})\times1000=2000+481.5\)
= 2481.5