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=55 and width of class interval, h=10.Table for given data is

Now, we have, N=90+x+y
\(\Rightarrow \quad 120=90+x+y\)
\(\Rightarrow \quad x+y=30\)
\(\because \quad Mean=59\)
\(\Rightarrow \quad a+\left( \frac { \sum { f_{ i }u_{ i } } }{ N } \right) \times h=59\)
\(\Rightarrow \quad 55+\left( \frac { 44+4y }{ 120 } \right) \times 10=59\)
\(\Rightarrow \quad \frac { 11+y }{ 3 } =4\)
\(\Rightarrow \quad 11+y=12\)
\(\Rightarrow \quad y=1\)
On putting the value of y in Eq.(i), we get
x + 1 = 30⇒ x = 29
Hence, x = 29 and y = 1.

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.

First ten prime numbers in ascending order are 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.
Here, n=10                      [even]
∴ Median
\(=\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { n }{ 2 } \right) th+\left( \frac { n }{ 2 } +1 \right) th \right] observation\\ =\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { 10 }{ 2 } \right) th+\left( \frac { 10 }{ 2 } +1 \right) th \right] observation\)
\(\)\(=\frac { 1 }{ 2 } \) [Value of 5th observation+Value of 6th ob
servation]
\(=\frac { 1 }{ 2 } [11+13]\)
\(=\frac { 24 }{ 2 }=2\)
\(\)

Given, median=137 units and mean=137.05 units
We know that,
Mode=3(Median-2(Mean)
=3(137)-2(137.05)
=411-274.10=136.90
Hence, the value of mode is 136.90 units.

167.13

Let frequency of class 20-30 be f₁ and that of class 40-50 be f_2.
Since, the sum of all frequency is 170.
The cumulative frequency table for given distribution is

Here, median=35
So, the median is 30-40. Also, n=170\(\Rightarrow \quad \frac { n }{ 2 } =85\)
∴ l=30, f=40, c.f=30+f₁ and h=10
Now median \(=l+\left\{ \frac { \frac { n }{ 2 } -cf }{ f } \right\} \times h\)
\(\Rightarrow \quad 35=30+\left\{ \frac { 85-(30+f_{ 1 }) }{ 40 } \right\} \times 10\\ \Rightarrow \quad 35\times 4=120+(55-f_{ 1 })\\ \Rightarrow \quad 140=175-f_{ 1 }\Rightarrow f_{ 1 }=35\)
Also, 110+f₁+f₂=170         [∵ sum of all frequencies=170, given]
\(\Rightarrow \quad f_{ 1 }+f_{ 2 }=60\\ \Rightarrow \quad 35+f_{ 2 }=60\\ \therefore \quad f_{ 2 }=60-35=25\)
Hence, the missing frequency of the class 20-30 is 35 and the class 40-50 is 25.

Mean = \(\frac { \Sigma f_{ i }x_{ i } }{ \Sigma f_{ i } } =\frac { 1640 }{ 50 } \times 32.8\)
Modal class = 30-40
l=30, f₁=15 , f₂=7, f₀=10 , h=10
Mode = \(l+\frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }f_{ 0 }-f_{ 2 } } \times h\)
\(30+\frac { 15+10 }{ 30-10-7 } \times 10\)
30+3.85
=33.85

Benefits of regular medical check-up are
(t) It enables us to take pre-medical action on time.
(it) It reduces the chances of getting sick.

Mean = \(\frac { \Sigma x_{ i }f_{ i } }{ \Sigma f_{ i } } =\frac { 11000 }{ 50 } \)
= 220
Average daily Income = Rs 220