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.

\(N=50\Rightarrow \frac { N }{ 2 } =\frac { 50 }{ 2 } =25\)
So ,Median class =120-140
Median = l + \(l+\left( \frac { \frac { N }{ 2 } -c.f }{ f } \right) \times h\)
  = \(120+\frac { 260 }{ 14 } \)
  = 120+18.57
Median = 138.57

x+y=60-45=15
Median = 28.5
Median class = 20 - 30
Median = l + \(\frac { \frac { N }{ 2 } c.f }{ f } \times h\)
\(28.5=20+\frac { \left\lfloor 30-(x+5) \right\rfloor }{ 20 } \times 10\)
\(8.5=\frac { 25-x }{ 2 } \)
25-x=17 x=25-17=8
y=15-8=7

Here Modal Class = 20-30
i=20,f1=25,,f2=13 , f0=12 , h=10
\(=l+\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \times h\)
\(20+\frac { 25-12 }{ 50-12-13 } \times 10\)
 =20+5.2=25.2

Let us make the following table for the given data.

\(\text { Here, } \Sigma f_{i}=30 \text { and } \Sigma f_{i} x_{i}=1860.0\)
On putting the values of \( \Sigma f_{i} x_{i}\) and \( \Sigma f_{i}\) in the formula
\(\bar{x}=\frac{\Sigma f_{i} x_{i}}{\Sigma f_{i}}\) we get
\(\bar {x} =\frac{1860.0}{30}=62\)
Hence, the mean marks obtained by the students are 62.

Given, radius of cone (r) = 7 cm
and height of cone (h) = 2r = 14 cm

\(\therefore\)  Volume of the solid = Volume of cone + Volume of hemisphere
\(=\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3\)
\(=\frac{1}{3} \times \frac{22}{7} \times 7^2 \times 14+\frac{2}{3} \times \frac{22}{7} \times(7)^3\)
\(=\frac{1}{3} \times \frac{22}{7} \times 7^3(2+2)=\frac{22}{3} \times 7^2 \times 4\)
\(=\frac{4312}{3} \mathrm{~cm}^3\)

The cumulative frequencies with their respective class intervals are as follows.

It can be observed that the cumulative frequency just greater than  n/2 (i.e 400/2 = 200) is 216
belonging to class interval 3000 − 3500.
Median class = 3000 − 3500
Lower limit (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size (h) = 500
\(\text { Median }=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \)
\(=3000+\left(\frac{200-130}{86}\right) \times 500 \)
\(=3000+\frac{70 \times 500}{86}\)
= 3406.976
Therefore, median life time of lamps is 3406.98 hours.

(i) 45.8 kg
(ii) The values like humanity, social service and honesty in performing duty are depicted here.

Here, a= 12, Now 12 + b = 25
\(\Rightarrow\)  b = 13
25 + 10 = c
\(\Rightarrow\)   c = 35
c + d = 43
\(\Rightarrow\)  35 + d = 43 \(\Rightarrow\) d = 8
43 + e = = 48 \(\Rightarrow\) e = 5, 48 + 2 = f
\(\Rightarrow\) f = 50
\(\therefore\)  a = 12, b = 13, c = 35,
d = 8, e = 5, f = 50
So given distribution becomes

Modal class is 155 - 160
l = 155,f₀ = 12,f₁ = 13,f₂ = 10,h = 5
Mode = l + \(\left( { {{f}_{1}-{f}_{0} }\over{ 2{f}_{1}-{f}_{0}-{f}_{2}} } \right)\times h\)
= 155 + \({ { 13-12 }\over{ 2\times13-12-10} }\times5\)
= 155 + \({ { 5 }\over{ 4 } }\) = 155 + 1.2 = 156.23

(i) The cumulative frequency table of given frequency distribution is

Here, n = 68
\(\therefore \quad \frac{n}{2}=34\)
Since, the cumulative frequency just greater than 34 is 42 and the corresponding class is 125- 145. So, the median class is 125-145.
Now, l = 125, f = 20, cf = 22 and h = 20
\(\begin{aligned} & \therefore \text { Median }=l+\left\{\frac{\frac{n}{2}-c f}{f}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} \quad=125+\left\{\frac{34-22}{20}\right\} \times 20=125+12=137 \text { units } \end{aligned}\)
(ii) Let the assumed mean, a = 135
and width of the class, h = 20
Table for the given data is

We have, N = 68, and \(\sum\)f_iu_i = 7
By step deviation method,
Mean = \(\begin{aligned} \text { Mean } & =a+b \times \frac{1}{N} \times \Sigma f_i u_i=135+20 \times \frac{1}{68} \times 7 \\ \end{aligned}\)
\(\begin{aligned} & =135+\frac{35}{17}=135+2.05=137.05 \text { units } \end{aligned}\)
(iii) Here, the modal class is 125-145 having maximum frequency, f₁ = 20.
Now, f₀ = 13, f₂ = 14, l = 125 and h = 20
\(\begin{aligned} \therefore \text { Mode } & =l+\left\{\frac{f_1-f_0}{2 f_1-f_0-f_2}\right\} \times h \\ \end{aligned}\)
\(\begin{aligned} & =125+\left\{\frac{20-13}{40-13-14}\right\} \times 20 \\ \end{aligned}\)
\(\begin{aligned} & =125+\frac{7 \times 20}{13}=125+\frac{140}{13} \end{aligned}\)
= 125 + 10.77 = 135.77 units
Hence, median = 137 units, mean = 137.05 units and mode = 135.77 units
So, we conclude that three measures are approximately the same.
(iv) Electricity consumption can be reduced by
(a) switching off electric appliances when not in use.
(b) using star rated AC's, refrigerators.

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
\(\therefore \quad \overline { x } =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h=20+\left\{ \frac { 30 }{ 40 } \right\} \times 10\\ =20+\frac { 30 }{ 4 } =\frac { 80+30 }{ 4 } =\frac { 110 }{ 4 } =27.5\)

\(\because \quad \overline { x } =\frac { \sum { f_{ i }x_{ i } } }{ n } \Rightarrow \sum { f_{ i }x_{ i } } =n\overline { x }\)
\( \\ \therefore \quad \sum { \left( f_{ i }x_{ i }-\overline { x } \right) } =\sum { f_{ i }x_{ i } } -\sum { \overline { x } }\)
\( \\ =n\overline { x } -n\overline { x }\)
\( \\ =0\)

45

So, frequency of the class 30 - 40 is 3.
Class mark of the class: 10-25 = \(\frac { 10+25 }{ 2 } \)
 \(=\frac { 35 }{ 2 } =17.5\)

Modal class = 35-40
l=35, f₁=50 , f₂=42 , f₀=34 , h=5
Mode = l+ \(\frac { (f_{ 1 }-f_{ 0 }) }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \times h\)
\(=35+\frac { 50-34 }{ 100-34-42 } \times 5\)
\(=35+\frac { 16\times 5 }{ 24 } =38.33\)

(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): \(n=51 . \mathrm{So}, \frac{n}{2}=\frac{51}{2}=25.5 .\)
This observation lies in the class 145 - 150
= 145
(ii) (b): 150
(iii) (a): 145
(iv) (c): 5
(v) (a): l ( lower limit) = 145, cf = 11, f = 18, h = 5.
\(\text { Median }=l+\left(\frac{\frac{n}{2}-\mathrm{cf}}{f}\right) \times h\)
\(\text { Median }=145+\left(\frac{25.5-11}{18}\right) \times 5=145+\frac{72.5}{18}\)
= 149.03 cm

(a) (ii) Since the highest frequency is 23 which belongs to 35 – 45.
Therefore, modal class is 35 – 45.
(b) (ii) Here, \(n=80 \Rightarrow \frac{n}{2}=40\) 
which lies in 35 – 45 
Therefore, medial class is 35 – 45.
(c) (iii)   Here,l = 35, f₀ = 21, f₁= 23, f₂ = 14, h = 10.
\(\text { Mode }=l+\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}} \times h \quad \Rightarrow \text { Mode }=35+\frac{23-21}{46-21-14} \times 10\) 
(d) (i) 35-45
(e) (ii)19