Chapter 15: Statistics

Ans: $\sum {{f_i} = 60} $

$\Rightarrow (x - 2) + x + {x^2} + {(x + 1)^2} + 2x + x + 1 = 60$

$\Rightarrow 2{x^2} + 7x - 60 = 0$

Solving the above equation, we get $x =  - \dfrac{{15}}{2},4$

$\therefore x = 4.$

Let the mean is 3. 

Mean: 

$\bar{x}=A+\left(\dfrac{\sum f_{i} d_{i}}{n}\right)$ $=3+\left(\dfrac{-12}{60}\right)=3-0.2=2.8$

So the mean is 2.8.

Standard deviation: 

$\sigma=\sqrt{\dfrac{\sum f_{i} d_{i}^{2}}{n}-\left(\dfrac{\sum f_{i} d_{i}}{n}\right)^{2}}$ 

$=\sqrt{\dfrac{78}{60}-\left(\dfrac{-12}{60}\right)^{2}}=\sqrt{1.3-\dfrac{144}{3600}}=\sqrt{1.26}=1.122$ 

So the Standard deviation $=1.122$ 

Ans: Given: Mean life of a sample of \[60\] bulbs was \[650\] hours and the standard deviation was \[8\] hours is $ 8$ hours.

With the help of the below formula,

\[\sigma    = \sqrt {\dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{\left( {\overline {{x_1}}  - \overline {{x_2}} } \right)}^2}}}{{{{\left( {{n_1} + {n_2}} \right)}^2}}}} \]

Then,

${n_1} = 60 ,  {{\bar x}_1} = 650 ,  {s_1} = 8; {n_2} = 80,{{\bar x}_2} = 660,  {s_2} = 7$

$\sigma  = \sqrt{\dfrac{{{n_1}{s_1}^2 + {n_2}{s_2}^2}}{{{n_1} + {n_2}}} + \dfrac{{{n_1}{n_2}{{(\overline {{x_1}}  - \overline {{x_2}} )}^2}}}{{{{({n_1} + {n_2})}^2}}}}$ 

$\Rightarrow \sigma= \sqrt {\dfrac{{60 \times {8^2} + 80 \times {7^2}}}{{60 + 80}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{{{\left( {60 + 80} \right)}^2}}}}$        

Since, $=\dfrac{{60 \times 64 + 80 \times 49}}{{140}} = 20\dfrac{{\left[ {3(64) + 4(49)} \right]}}{{140}}$

$=  \dfrac{{\left[ {3(64) + 4(49)} \right]}}{7} =  \dfrac{{192 + 196}}{7}$ 

$\sigma= \sqrt {\dfrac{{60 \times 64 + 80 \times 49}}{{140}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{{{140}^2}}}}$

$\Rightarrow \sigma= \sqrt {\dfrac{{60 \times 64 + 80 \times 49}}{{140}} + \dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{140 \times 140}}}$

$\dfrac{{60 \times 80{{\left( {650 - 660} \right)}^2}}}{{140 \times 140}} = \dfrac{{60 \times 80 \times 100}}{{140 \times 140}}$

$= \dfrac{{1200}}{{49}}$

$\Rightarrow \sigma   =  \sqrt {\dfrac{{192 + 196}}{7} + \dfrac{{1200}}{{49}}}  = \sqrt {\dfrac{{2716 + 1200}}{{49}}}$

$= \sqrt {\dfrac{{3916}}{{49}}}  = \dfrac{{62.58}}{7} = 8.9$

Ans: Given:

(a)  Mean 

(b)  The standard deviation of respectively.

With the help of the below formula,

\[{\sigma ^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)^2}\]

$\bar x = 50,n = 100 {\text{and}} \sigma  = 4$

$\dfrac{{\sum {x_i}}}{{100}} = 50$

$\sum\limits_{}^{} {{x_i} = 5000} $

${\sigma ^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {\dfrac{{\sum {f_i}{x_i}}}{{\sum {f_i}}}} \right)^2}$ 

\[\Rightarrow {4^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} - {\left( {50} \right)^2}\]

$\Rightarrow {4^2} + {\left( {50} \right)^2} = \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}}$

$\Rightarrow \dfrac{{\sum {f_i}{x_i}^2}}{{\sum {f_i}}} = 16 + 2500$

$\Rightarrow \dfrac{{\sum {f_i}{x_i}^2}}{{100}} = 2516$

$\Rightarrow \sum {f_i}{x_i}^2 = 251600$ 

Ans: Given: \[\sum\left( {x - 5} \right) = 3 {\text{ and }} {\sum\left( {x - 5} \right)^2} = 43\]

The total number of items is \[18\] .

Mean   \[ = {\text{A}} + \dfrac{{\sum \left( {x - 5} \right)}}{n}\]

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {x - 5} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {x - 5} \right)}}{n}} \right)}^2}} \].

Then,

\[\sum \left( {x - 5} \right) = 3 {\text{ and }} {\sum \left( {x - 5} \right)^2} = 43\]

Mean \[ = {\text{A}} + \dfrac{{\sum \left( {x - 5} \right)}}{n}\]

$ = 5 + \dfrac{3}{{18}}$

$= 5 + 0.1666$

$= 5.17$ 

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {{\left( {x - 5} \right)}^2}}}{n} - {{\left( {\dfrac{{\sum \left( {x - 5} \right)}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{43}}{{18}} - {{\left( {\dfrac{3}{{18}}} \right)}^2}} $

$= \sqrt {2.3889 - 0.0277}  = 1.53$ 

Ans: By using the below basic formulas, we can find the mean and variance.

Mean \[ = \dfrac{{\sum {f_i}{y_i}}}{{{f_i}}}\]

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {f_i}{y_i}^2}}{n} - {{\left( {\dfrac{{\sum {f_i}{y_i}}}{n}} \right)}^2}} \]

Mean  \[ = \dfrac{{\sum {f_i}{y_i}}}{{{f_i}}}\]

$= \dfrac{{66.5}}{{16}}$

$= 4.16$ 

Standard deviation   \[\sigma  = \sqrt {\dfrac{{\sum {f_i}{y_i}^2}}{n} - {{\left( {\dfrac{{\sum {f_i}{y_i}}}{n}} \right)}^2}} \]

$= \sqrt {\dfrac{{340.25}}{{16}} - {{\left( {\dfrac{{66.5}}{{16}}} \right)}^2}} $

$= \sqrt {21.26 - 17.22}  = 4.04$