Median = \(\frac { 1 }{ 3 } \) Mode+ \(\frac { 2 }{ 3 } \) Mean
= \(\frac { 1 }{ 3 } (12.4)+\frac { 2 }{ 3 } (10.5)\)
\(=\frac { 12.4 }{ 3 } +\frac { 21 }{ 3 } \)
\(=\frac { 12.4+21 }{ 3 } =\frac { 33.4 }{ 3 } \)
\(\frac { 33.4 }{ 3 } =11.13\)

The cumulative frequency table for given data is

Here, \(\frac { n }{ 2 } =\frac { 67 }{ 2 } =33.5\)
The cumulative frequency just greater than 33.5 is 42 and the corresponding class is 125-145.
Thus, we have median class 125-145.
Also, the maximum frequency is of the class 125-145.
Therefore, the modal class is 125-145.
Difference of the upper limit of median class and the lower limit of modal class=145-125=20

Here, the highest frequency is 30, which lies in the interval 30-40. So, it is the modal class, i.e. 30-40.

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

Probability of an event cannot be negative

Area of the circle =\(\pi { r }^{ 2 }\)

=\(\frac {22} {7} \times 210 \times 210\)
= 138600km²
Area of shaded region = 60 x 30
= 1800 km² 
Required probability = \(\frac {1800} {138600} =\frac {1} {77}\) 
(i) (a) Report your parents first. 

Total balls in a bag = 14
Number of white balls = x
Now, 6 more white balls are added to the bag.
\(\Rightarrow \)Number of white balls = x + 6
And, total balls in the bag = 14 + 6 = 20
Now, according to the statement of the question 
\(\frac { x+6 }{ 20 } =\frac { 1 }{ 2 } \)
\(\Rightarrow \) 2x+12=20
\(\Rightarrow \) 2x=8
\(\Rightarrow \) x=4 

Let a = 28, h = 5

\(\bar{x}=28-{ {5541 }\over{9350 } }\times5\) 
= 28 - 2.96 = 25.04

\(\overset { - }{ x } =a+\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
\(=35+\frac { -44 }{ 65 } \times 10=35-6.76=28.24\)

Total cards = 49 
(i) P(odd number) = \(\frac{25}{49}\)
(ii) P(multiple of 5) = \(\frac{9}{49}\)
(iii) P(even prime) = \(\frac{1}{49}\)

Here, the given data is in ascending order of x_i. 
Cumulative frequency table for the given data is

Here, n=120 [even]
Median\(=\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { n }{ 2 } \right) th+\left( \frac { n }{ 2 } +1 \right) th \right] observation\)
\(\\ =\frac { 1 }{ 2 } \times Value\quad of\quad \left[ \left( \frac { 120 }{ 2 } \right) th+\left( \frac { 120 }{ 2 } +1 \right) th \right] observation\)
\(=\frac { 1 }{ 2 } \)[Value of 60th observation+Value of 61th observation]
Both 60th and 61th observations lie in the cumulative frequency 65 and its corresponding value of x is 5.
Median\(\) \(=\frac { 1 }{ 2 } [5+5]=5\)

Table for given data is

Here, \(\sum { f_{ i } } =39+p\) and \(\sum { f_{ i }x_{ i }=2370+30\quad p } \)
Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } \)
\(54=\frac { 2370+30p }{ 39+p } \quad \left[ \because \quad mean=54,\quad given \right] \)
\(\Rightarrow \quad 54(39+p)=2370+30p\\ \Rightarrow \quad 2106+54p=2370+30p\\ \Rightarrow \quad 24p=264\\ \Rightarrow \quad p=11\)
Hence, the value of p is 11.

(i) Total number of possible outcomes=4
∴ n(S)=4
Let E be the event of getting correct answer,
∴ Total number of favourable outcomes,
 n(E)=1
P(getting his correct answer)\(=\frac { n(E) }{ n(S) } =\frac { 1 }{ 4 } \)
(ii) If Shivam copies the answer, he will be violating the value of honesty.
(iii) Cheating may get him marks in this test but this habit may not let him develop an integrity of character in the long run.

Let the number of green marbles out of 24 marbles in a jar be x.
Probability of getting green marbles = \(\frac { 2 }{ 3 } \)
\(\frac { x }{ 24 } =\frac { 2 }{ 3 } \)
3x = 48
x = 16
 Number of green marbles = 16
Hence, the number of blue marbles = 24 - 16 
=8

Ans. 11

Mode = \(l+\frac { f-{ f }_{ 0 } }{ 2f-{ f }_{ 0 }-{ f }_{ 2 } } \times h\)
\(65=60+\frac { 12-{ f }_{ 1 } }{ 2\times 12-{ f }_{ 1 }-6 } \times 20\)
\(5=\frac { 12-{ f }_{ 1 } }{ 18-{ f }_{ 1 } } \times 20\)
90 - 5f₁ = 240 - 20 f₁
15f₁ = 150 ⇒ f₁ =10.
∵ Mode = 65
∴ Modal class = 60 - 80 as its frequency is 12.
I = 60, f = 12, f_o= f₁, f₂= 6, h = 20.

Let a = Assumed mean = 54
\(\overset { - }{ x } =a+\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
Mean = 54+\(\frac { 18 }{ 60 } \times 4=55.2\)
Maximum frequency =14 Modal class =52-56,l=52 f1=14,f0=10,f2=10,h=4
Mode=52+\(\frac { 14-10 }{ 28-10-10 } \times 4=54\)

Here, class interval are not continuous. But it does not affect mid-values. So, we will solve it without making it continuous.

Mean \(\left( \overline { x } \right) =\frac { \sum { f_{ i }x_{ i } } }{ \sum { f_{ i } } } =\frac { 362 }{ 28 } =12.93\)

No, the outcomes are not equally likely. The outcomes '3' is more likely to occur than the others.

Given, a group consists 12 persons.
\(\therefore \)  Total number of outcomes = 12
(i) Given, number of extremely patient persons = 3
\(\therefore \)  Number of favourable outcomes = 3
\(\therefore \)  P (extremely patient) = \(\frac{3}{12} = \frac{1}{4}\)
(ii) Given, number of extremely honest persons = 6 and number of extremely kind  persons
= 12 - 6 - 3 = 3
\(\therefore \)  Number of favourable outcomes = Number of extremely kind persons + Number of extremely honest persons = 6 + 3 = 9
\(\therefore \)  P (extremely kind or honest) = \(\frac{9}{12}=\frac{3}{4}\)