Mean= \(\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } \)
\(=\frac { 2755 }{ 70 } =39.36\)

Mean \(\overset { - }{ x } =\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \)
\(\Rightarrow 53=\frac { 3340+70p }{ 72+p } \)
\(\Rightarrow 3340+70p=53(72+p)\)
\(\Rightarrow 3340+70p=3816-3340\)
\(\Rightarrow 70p-53p=3816-3340\)
\(\Rightarrow 17p=476\)
\(p=\frac { 476 }{ 17 } =28\)

\(N=600\Rightarrow \frac { N }{ 2 } =300\)
Median class= 1000-2000
\(Median\quad =\quad l+\left( \frac { \frac { N }{ 2 } -c.f }{ f } \right) \times h\)
\(Median\quad =\quad 1000+\left( \frac { 300-250 }{ 190 } \right) \times 1000\)
\(=1000+\frac { 50 }{ 190 } \times 1000\)
=1000+263.16=1263.16
Median = Rs 1263.16

I term + 1
II term + 2
III term + 3
n terms + n
The mean of the new numbers is  \(\bar { x } +\frac { \frac { n\left( n+1 \right) }{ 2 } }{ n } \)
\(=​​\bar { x } +\frac { n\left( n+1 \right) }{ 2 } \)

To prove \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) =0 } \) / algebraic sum of deviation from mean is zero
We have,  \(\bar { x } =\frac { 1 }{ n } \left( \sum _{ i=1 }^{ n }{ { x }_{ i } } \right) \)
\(n\bar { x } =\sum _{ i=1 }^{ n }{ { x }_{ i } } \)
Now,  \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\left( { x }_{ 1 }-\bar { x } \right) +\left( { x }_{ 2 }-\bar { x } \right) +.........+\left( { x }_{ n }-\bar { x } \right) \)
\(\Rightarrow \quad \ \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\left( { x }_{ 1 }+{ x }_{ 2 }+.........+{ x }_{ n } \right) -n\bar { x } \)
\(\Rightarrow \quad \ \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =\sum _{ i=1 }^{ n }{ { x }_{ i } } -n\bar { x } \)
\(\Rightarrow \quad \sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =n\bar { x } -n\bar { x } =0\)
Hence,  \(\sum _{ i=1 }^{ n }{ \left( { x }_{ i }-\bar { x } \right) } =0\)

From table N=60+f=\(\frac { N }{ 2 } =\frac { 60+f }{ 2 } \)
since Median=240
Median class=200-300
Median = l+\(\left( \frac { \frac { N }{ 2 } -c.f }{ f } \right) \times h\)
\(240+200+\left( \frac { \frac { 60+f }{ 2 } -32 }{ f } \right) \times 100\)
\(\Rightarrow 40=\left( \frac { 60+f-64 }{ 2f } \right) \times 100\)
\(\Rightarrow 8f=10f-40\)
\(\Rightarrow 2f=40\)
 f=20

\(\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } =\frac { 2640 }{ 100 } =26.4\)
Median class=20-30
Median =20+ \(\frac { 50-24 }{ 36 } \times 10\)
=20+7.22=27.22

Mean = A+ \(\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
= 45 + \(45+\left( \frac { -37 }{ 150 } \right) \times 10\)
= 42.5 approx

Mean x=\(\frac { \Sigma x_{ i }f_{ i } }{ \Sigma f_{ i } } =\frac { 3020 }{ 50 } =60.4\)

We construct a cumulative frequency distribution of the less than type as follows:

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

Here, given mode, is 54, which lies between 45-60. Therefore, the modal class is 45-60.
l=45, f₁=16, f₀=f, f₂=12 and h=15
Mode \(=l+\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\)
\(54-45+\frac { 16-f }{ 32-f-12 } \times 15\Rightarrow 9=\frac { 16-f }{ 20-f } \times 15\\ \Rightarrow \quad 9(20-f)=15(16-f)\\ \Rightarrow \quad 180-9f=240-15f\\ \Rightarrow \quad 6f=240-180=60\Rightarrow f=10\)
Hence, required value of f is 10.

Here, maximum frequency is 315 and the class corresponding to this frequency is 20-40.

l=20, f₁=315, f₀=35, f₂=120 and h=20
Now, mode\(=l+\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\)
\(=20+\frac { 315-35 }{ 2\times 315-35-120 } \times 20\\ =20+\frac { 280 }{ 475 } \times 20=20+11.79=31.79\)
Hence, average age of maximum number of patients is 31.79.

Here, class intervals are not continuous. But mid-value x_i of each class interval would be same either class interval is continuous or not continuous.
So, we solve it without making it continuous.
Also, x_i are larger so we apply step-deviation method. Her, class width (h)=5. Table for the given data is

We have, a = 72, h = 5, \(\sum { f_{ i }=40 } \) and \(\sum { f_{ i }u_{ i } } =72\)
By step-deviation method,
Mean
\(\left( \overline { x } \right) =a+\frac { \sum { f_{ i } } }{ \sum { f_{ i }u_{ i } } } \times h=72+\frac { 22 }{ 40 } \times 5=72+2.75=74.75\)

Here, class width, h = 20 - 10- = 10. Nw, let the assumed
Mean, a = 35.
Then, table for the given data is

Now, we have,
\(\sum { f_{ i }u_{ i } } =66\), \(\sum { f_{ i } } =100\), a=35 and h=10
Mean \((\overline { x } )=a+\left( \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right) \times h\)
\(=35+\frac { 66 }{ 100 } \times 10\)
\(=35+\frac { 66 }{ 100 } \)
\(=35+6.6=41.6\)​​​​​​

Let the assumed mean, a=17. Now, let us make the following table.

Here, \(\sum { f_{ i } } =40\) and \(\sum { f_{ i }d_{ i } } =72\)
Mean \(\left( \overline { x } \right) =a+\frac { \sum { f_{ i }d_{ i } } }{ \sum { f_{ i } } } \)
\(=17+\frac { 72 }{ 40 } =17+1.8\)
\(=18.8\)