COLUMN - I		COLUMN - II
1)	If a die is rolled, then find the variance and standard deviation of the possibilities	p)	4.9
2)	The standard deviation of the average temperatures recorded over a five-day period last winter: 18, 22, 19, 25, 12 (The mean = 19.2)	q)	8
3)	Six numbers from list of nine integers are 7, 8, 3, 5, 9, and 5. What is the largest possible value of the median of all nine numbers in this list?	r)	109
4)	The median of a list of 99 consecutive integers is 60. What is the greatest integer in the list?	s)	1.708

When a die is rolled, the possible number of outcomes is 6. So the sample space, n = 6 and the data set = {1; 2; 3; 4; 5; 6}. To find the variance, first, we need to calculate the mean of the data set.

Mean, x = (1+2+3+4+5+6)/6=3.5

We can put the value of data and mean in the formula to get; ${\mathrm{\sigma }}^2=\sum (\mathrm{xi}-\overline{\mathrm{x}})2/\mathrm{n}$

$\begin{array}{l}{\mathrm{\sigma }}^2=1/6(6.25+2.25+0.25+0.25+2.25+6.25)\\ {\mathrm{\sigma }}^2=2.917\end{array}$

Therefore the variance is ${\mathrm{\sigma }}^2=2.917$, and standard deviation, $\mathrm{\sigma }=\sqrt{2.917}=1.708$

This time we will use a table for our calculations.

Mean = 19.2

To find the variance, we divide $5-1=4$

$\frac{94.8}{4}=23.7$

Finally, we find the square root of this variance.

$\sqrt{23.7}=4.9$

So the standard deviation for the temperatures recorded is 4.9; the variance is 23.7

Finally, we find the square root of this variance.

$\sqrt{23.7}\approx 4.9$

So, the standard deviation for the temperatures recorded is 4.9; the variance is 23.7.

$\Rightarrow$The given six numbers are 7, 8, 3, 5, 9 and 5

$\Rightarrow$Arrange the six numbers in order 3, 5, 5, 7, 8, 9

$\Rightarrow$We have three more numbers to insert into the list, and the median will be the highest 5 (and 5 lowest) number on the list. If the three numbers are greater than 9, the median will be the highest it can possible be.

$\therefore$The median is the 5th item of data which is 8

$\therefore$Median = 8

51. Correct option is A

Given that:

Median of 99 consecutive integers = 60

Median of 99 consecutive integers = 60

$\text{ i.e. }{\mathrm{t}}_{50}=60$

Let a be first term and common difference =1 ( as integers are consecutive)

${\mathrm{t}}_{50}=60$

$\text{ or, }\mathrm{a}+49\times 1=60\text{ or, }\mathrm{a}=11$

Greatest integers $\text{ i.e. }{\mathrm{t}}_{99}=\mathrm{a}+98\times 1=11+98=109$

Hence, A is the correct option.