Statistics

$\text { Given, } \text { mean }=6 \text { and variance }=\frac{9}{4}$

$\therefore \frac{4+5+6+6+7+8+x+y}{8}=6$

$ \begin{array}{lrl} \Rightarrow & 36+x+y=48 \\ \Rightarrow & x+y=12 \quad \text{... (i)} \end{array}$

And

$\frac{1}{8}\left(4^2+5^2+6^2+6^2+7^2+8^2+x^2+y^2\right)-6^2=\frac{9}{4}$

$ \begin{array}{rrrr} \Rightarrow & \frac{1}{8}\left(226+x^2+y^2\right) =\frac{153}{4} \\ \Rightarrow & x^2+y^2 =80 \quad \text{... (ii)} \end{array} $

On solving Eqs. (i) and (ii), we get $ \begin{aligned} x & =4 \text { and } y=8 \quad [\because x < y]\\ \therefore \quad x^2-2 y & =(4)^2-2(8) \\ & =16-16=0 \end{aligned} $

Let the two unknown items be x and y. Then,

Mean $ = 4 \Rightarrow {{1 + 2 + 6 + x + y} \over 5} = 4$

$\Rightarrow$ x + y = 11 ..... (i)

and variance = 5.2

$ \Rightarrow {{{1^2} + {2^2} + {6^2} + {x^2} + {y^2}} \over 5}$ $-$ (mean)² = 5.2

$\Rightarrow$ 41 + x² + y² = 5(5.2 + 16)

$\Rightarrow$ 41 + x² + y² = 106

$\Rightarrow$ x² + y² = 65 ..... (ii)

On solving Eqs. (i) and (ii), we get

x = 4, y = 7 or x = 7, y = 4

Given, E(X) = 3 and E(X²) = 11

Variable of X = E(X²) $-$ [E(X)]² = 11 $-$ 3² = 11 $-$ 9 = 2

Given, x₁ + x₂ + x₃ + .... + x₁₀ = 12

and x$_1^2$ + x$_2^2$ + ..... + x$_{10}^2$ = 18

${\sigma ^2} = {1 \over n}\sum {{x^2} - {{\left( {{1 \over n}\sum x } \right)}^2}} $

$ = {1 \over {10}} \times 18 - {\left( {{1 \over {10}} \times 12} \right)^2} = {9 \over 5} - {{36} \over {25}} = {9 \over {25}}$

$\therefore$ Standard deviation = 3/5.

Let us make the table from the given data

For player A, Mean = ${{\sum {{x_i}} } \over n} = {{610} \over {10}} = 61$

$SD = \sqrt {{{\sum {{{({x_i} - 61)}^2}} } \over N}} $

$ = \sqrt {{{526} \over {10}}} = 7.25$

For player B, Mean = ${{\sum {{y_i}} } \over n} = {{610} \over {10}} = 61$

$SD = \sqrt {{{\sum {{{({y_i} - 61)}^2}} } \over N}} $

$ = \sqrt {{{11416} \over {10}}} = 33.79$

Since, SD for player A is 7.25 < SD for player B is 33.79.

Hence, player A is more consistent player.

Let two observations be x and y.

$\therefore$ Mean $ = {{1 + 3 + 8 + x + y} \over 5} = {{12 + x + y} \over 5}$

$ \Rightarrow 5 = {{x + y + 12} \over {15}} \Rightarrow x + y = 13$ ..... (i)

Now, variance $ = {{\sum {x_i^2} } \over n} - {(\overline x )^2}$

$ \Rightarrow 9.20 = {{{1^2} + {3^2} + {8^2} + {x^2} + {y^2}} \over 5} - 25$

$ \Rightarrow 46 = 74 + {x^2} + {y^2} - 125$

$ \Rightarrow {x^2} + {y^2} = 97$ ...... (ii)

From Eqs. (i) and (ii), we get

${(x + y)^2} - 2xy = 97$

$ \Rightarrow {(13)^2} - 2xy = 97$

$ \Rightarrow 2xy = 169 - 97 = 72$

$\Rightarrow$ xy = 36

$\therefore$ x : y = 4 : 9