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

The cumulative frequency for the given data is calculated as follows

From the table, it can be observed that n = 60
45 + x + y = 60
x + y = 15 (1)
Median of the data is given as 28.5 which lies in interval 20 - 30.
Therefore, median class = 20 - 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
\(\text { Median }=l+\left(\frac{\left(\frac{n}{2}\right)-c f}{f}\right) \times h\)
28.5 = 20 + [(60/2-(5+x))/20]xx10
8.5 = ((25-x)/2)
17 = 25 - x
8 + y =  15
y = 7
Hence, the values of x and y are 8 and 7 respectively.

Here, the given data is not in continuous classes, so we convert it intocontinuous classes for finding, the median. For this, change the classes to 117.5-126.5, 126.5-135.5, ..., 171.5-180.5 i.e., subtract 0.5 from lower limit of each class interval and add 0.5 to upper limit of each class interval.
Then, cumulative frequency table for given distribution is

Here, \(n=40 \Rightarrow \frac{n}{2}=\frac{40}{2}=20\)
the cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5. thus, we have the median class 144.5-153.5.
The cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5. thus, we have the median class 144.5-153.5.
Now, l = 144.5, f = 12, cf = 17, h = 9.
\(\begin{aligned} \therefore \text { Median } & =l+\left\{\frac{\frac{n}{2}-c f}{f}\right\} \times h=144.5+\left\{\frac{20-17}{12}\right\} \times 9 \\ \end{aligned}\)
\(\begin{aligned} =144.5+\frac{9}{4}=144.5+2.25=146.75 \mathrm{~mm} \end{aligned}\)

Given, \( \sum { f_{ i }u_{ i }=30 } \), \(\sum { f_{ i }=40 } \) and \(u_{ i }=\frac { x_{ i }-20 }{ 10 } \)
We know that, \(u_{ i }=\frac { x_{ i }-a }{ h } \)
On comparing, we get a=20, h=10
\(\therefore \quad \overline { x } =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h=20+\left\{ \frac { 30 }{ 40 } \right\} \times 10\\ =20+\frac { 30 }{ 4 } =\frac { 80+30 }{ 4 } =\frac { 110 }{ 4 } =27.5\)

Table for given distribution is

Here, n=\(\sum { f_{ i }=11 } \), \(\sum { f_{ i }x_{ i }=2a+52 } \)
Mean\(=\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } \)
\(\Rightarrow \quad \frac { 2a+52 }{ 11 } =6\)
\(\\ \Rightarrow \quad 2a+52=66\Rightarrow 2a=66-52=14\)
\(\\ a=\frac { 14 }{ 2 } =7\)

45

10

Modal class is 30 - 35, f = 30,f₁ = 25, f₀ = 10,f₂ = 7, h =5
Mode = l+ \(\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\Rightarrow \)Mode = 30+ \(\frac { 25-10 }{ 50-10-7 } \times 5\)
= 30+2.27 or 32.27 approx.

Median class : 40 - 50 \(\Rightarrow \) Lower Limit =40
Modal class   : 40 - 50  \(\Rightarrow \) Upper Limit = 50
Their sum = 40+50=90

(i) (b): Since, the highest frequency is 70, therefore the maximum number of children are of the age-group 10-12.
(ii) (a): Since, the modal class is 10-12
\(\therefore\) Lower limit of modal class = 10
(iii) (c) : Here,f₀ = 58,f₁ = 70 and f₂ = 42
Thus, the frequency of the class succeeding the modal class is 42.
(iv) (d): Mode \(=l+\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right] \times h\)
\(=10+\left[\frac{70-58}{140-58-42}\right] \times 2\)
\(=10+\frac{12}{40} \times 2=10+\frac{24}{40}=10.6 \text { years }\)
(v) (c): Given that, Mean = Mode
\(\therefore\) By Empirical relation, we have
Mode = 3 Median - 2 Mean
\(\Rightarrow\) Mode = 3 Median - 2 Mode
\(\Rightarrow\) 3 Mode = 3 Median
\(\Rightarrow\) Median = Mode = Mean

(i) (a): 

Here, \(\Sigma f_{i}=n=68 \text { then } \frac{n}{2}=\frac{68}{2}=34\) which lies in interval 125 - 145
= 125
(ii) (a): 125
(iii) (b): 145
(iv) (c): 20
(v) (a): Median class = 125 - 145
So, l = 125; n = 68; f = 20; cf = 22 and h = 20
Using formula, Median \(=l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h\)
\(=125+\left\{\frac{\frac{68}{2}-22}{20}\right\} \times 20\)
\(=125+\frac{34-22}{20} \times 20=125+12\)
= 137