Solution:

Formula used 

Here, n = 13

$$\sqrt{\frac{\left(\left(\ 13\right)^2\ -\ 1\right)}{12}}\ =\ \sqrt{\ \frac{\left(13+1\right)\left(13-1\right)}{12}}\ =\ \sqrt{\ \frac{14.12}{12}}\ =\ \sqrt{\ 14}$$

$$\sigma\ =\ \sqrt{\ 14}\ $$ Answer

Solution:

Relation used

$$3median=mode+2mean$$

3 median - 2 mean = mode

3 x 48 - 2 x 56 = 32 Answer

Solution:

The mean of given set is =$$\frac{\left(15+17+20+23+25\right)}{5}$$ = 20

mean deviation of the observations about the arithmetic mean = $$=\frac{\left(5+3+0+3+5\right)}{5}\ =\ \frac{16}{5\ }=3.2$$

3.2 Answer

Solution:

Nearest multiple of 11 before 150 = 143

So total 13 nos. below 150 are multiples of 11.

Sample space = 150

Probability = $$\frac{13}{150}$$ Answer

If a 4 digit number is formed at random without repetition using the digits 0, 1, 2, 3, 4, then the probability that the number thus formed is even is

There are 4 places: _ _ _ _

As repetition is not allowed,

Thousands place cannot have 0, so there are only 4 choices for that position.

Hundreds place can have all other remaining 4 choices for that position.

Tens place  can have all other remaining 3 choices for that position.

Units places can have only remaining 2 choices

So, through fundamental process of counting

No. of numbers possible = 4 x 4 x3 x 2 =96

For even nos. units place should be either 2 or 4 or 0

In a similar way of counting lets first fix 2 or 4 

So for thousands place 3 options are remaining (except 0).

Hundreds place can have all other remaining 3 choices for that position.

Tens place can have all other remaining 2 choices for that position.

Units places can have only remaining 2 choices. (2 or 4)

No. of numbers possible = 3 x3 x 2 x 2 = 36

In a similar way of counting lets first fix 0

So for thousands place 4 options are remaining (except 0).

Hundreds place can have all other remaining 3 choices for that position.

Tens place can have all other remaining 2 choices for that position.

Units places can have only remaining 1 choices. (only 0)

No. of numbers possible = 4 x 3 x 2 x 1= 24

Total Even nos. 36+24 = 60

Probability = 60/96 = 5/8 Answer

Solution:For mode of tabulated data, the observation with highest frequency is the mode.

Here maximum frequency is 24, the correpsonding observation is 8.

Answer 8

Data Set: 28, 4, 21, 42, 10, 9, 24, 30, 16, 5, 13

To find median, we first need to sort them in ascending order,

4,5,9,10,13,16,21,24,28,30,42

No, of terms,n= 11,

So median = $$\left(\ \frac{n+1}{2}\right)\ term\ if\ n\ is\ odd$$

= $$\left(\ \frac{11+1}{2}\right)\ =\ \frac{12}{2\ }\ =\ 6th\ term$$

= 16 Answer

Option (C)

Arithmtic Mean = $$\frac{\Sigma\ \left(\ distinct\ values\ \times\ frequency\right)}{\Sigma\ \left(frequency\right)}\ $$

                       =  $$\frac{6\ \times\ 2\ \ +\ 5\times\ 5\ +\ 4\times\ 3\ +\ 3\times\ 4}{2+5+3+4}\ $$

                       = $$\frac{61}{14}$$ Answer