Q: Find the range and coefficient of range of the following data. \$59, 46, 30, 23, 27, 40, 52, 35, 29\$

\$\Rightarrow\$ The given numbers are \$59,46,30,23,27,40,52,35,29\$.\$\Rightarrow\$ Here, the largest value \$=x_m=59\$ and the smallest value \$=x_0=23\$.\$\Rightarrow\$ \$Range=x_m-x_0\$ \$=59-23\$ \$=36\$\$\Rightarrow\$ \$Coefficient\,of\,range=\dfrac{x_m-x_0}{x_m+x_0}\$ \$=\dfrac{59-23}{59+23}\$ \$=\dfrac{36}{82}\$ \$=0.44\$

Q: The following observations have been arranged in ascending order. If the median of the data is \$63\$, find the value of \$x\$.\$29, 32, 48, 50, x, x+2, 72, 78, 84, 95\$

Total observations \$= 10\$ (even)Median \$= \dfrac{10}{2} = 5\$th & \$\dfrac{10}{2} + 1 = 6^{th}\$ observationsMedian \$= \dfrac{5^{th} + 6^{th} \text{ observations}}{2}\$\$63 = \dfrac{x+x+2}{2}\$\$\Rightarrow x+1 = 63\$\$\Rightarrow x=62\$

Q: The standard deviation of \$1, 2, 3, 4, 5, 6, 7\$ is?

Given observation is \$1,2,3,4,6,7\$Mean=\$\cfrac { 1+2+3+4+5+6+7 }{ 7 } =\cfrac { 28 }{ 7 } =4\$Variance=\$\sum _{ +r=1 }^{ n }{ \cfrac { \left( x-{ x }_{ i } \right) ^{ 2 } }{ n } } =\cfrac { \left( 1-4 \right) ^{ 2 }+\left( 2-4 \right) ^{ 2 }+\left( 3-4 \right) ^{ 2 }+\left( 4-4 \right) ^{ 2 }+\left( 5-4 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 }+\left( 7-4 \right) ^{ 2 } }{ 7 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 }+0+{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 7 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 9+4+1+1+4+9 }{ 7 } =\cfrac { 28 }{ 7 } =4\$standard deviation=\$\sqrt { variance } \\ \sqrt { 4 } =2\$

Q: If the mean of the observations:\$x, x + 3, x + 5, x + 7, x + 10\$ is \$9\$, the mean of the last three observations is

Mean of \$x, x + 3, x + 5, x + 7, x + 10 = 9\$Sum of the terms \$= 5x + 25\$Mean = \$\dfrac{5x+25}{5} =9\$\$x +5 =9\$\$x =4\$Last three observations are \$9,11,14\$Mean \$= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}\$

Q: Find the mean and variance for the following frequency distrubutionClasses0-1010-2020-3030-4040-50Frequencies5815166

Class interval\$ f_i\$\$x_i\$ \$u_i=\cfrac{x_i-A}{h}\$ \$u_i^2\$ \$f_iu_i\$ \$f_iu_i^2\$ 0-10 5 5 -2 4 -10 20 10-20 8 15 -1 1 -8 8 20-30 1525 0 0 0 0 30-40 1635 1 1 16 16 40-50 6 452 4 12 24 Total 50 10 68We have \$N=\sum f_i=50\$ Width of class \$h= 10\$Assumed mean \$A=25 (say)\$Mean, \$ \displaystyle \bar{x}=A+\frac{\sum_{i=1}^{5}f_{i}u_{i}}{N}\times h\$\$=25+\cfrac{10}{50}\times 10=25+2=27 \$\$\Rightarrow \bar{x}=27\$Variance \$\sigma ^{ 2 }=h^{ 2 }\left[ \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i }^{ 2 } }{ N } -\left( \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i } }{ N } \right) ^{ 2 } \right] \$\$\sigma ^{ 2 }=10^{ 2 }\left[ \cfrac { 68 }{ 50 } -\left( \cfrac { 10 }{ 50 } \right) ^{ 2 } \right] \$\$=10^{ 2 }\cfrac { \left[ 50\times 68-(10)^{ 2 } \right] }{ \left( 50 \right) ^{ 2 } } \$\$ =\displaystyle \frac{1}{25}\left [ 3400-100 \right ]\$\$=\cfrac{3300}{25} \$\$=132\$

Question 1

Find the range and coefficient of range of the following data. $59, 46, 30, 23, 27, 40, 52, 35, 29$

Accepted Answer

$\Rightarrow$  The given numbers are $59,46,30,23,27,40,52,35,29$.$\Rightarrow$  Here, the largest value $=x_m=59$ and the smallest value $=x_0=23$.$\Rightarrow$  $Range=x_m-x_0$                   $=59-23$                   $=36$$\Rightarrow$  $Coefficient\,of\,range=\dfrac{x_m-x_0}{x_m+x_0}$                                                $=\dfrac{59-23}{59+23}$                                                $=\dfrac{36}{82}$                                                $=0.44$

Question 2

The following observations have been arranged in ascending order. If the median of the data is $63$, find the value of $x$.$29, 32, 48, 50, x, x+2, 72, 78, 84, 95$

Accepted Answer

Total observations $= 10$ (even)Median $= \dfrac{10}{2} = 5$th & $\dfrac{10}{2} + 1 = 6^{th}$ observationsMedian $= \dfrac{5^{th} + 6^{th} \text{ observations}}{2}$$63 = \dfrac{x+x+2}{2}$$\Rightarrow x+1 = 63$$\Rightarrow x=62$

Question 3

The mean and standard deviation of $20$ observation are found to be $10$ and $2$ respectively. On rechecking it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases :(i) If wrong item is omitted(ii) If it is replaced by $12$

Accepted Answer

(i)  Given, number of observations $n=20$Incorrect mean $= 10$Incorrect standard deviation $= 2$$ \displaystyle \bar{x}=\frac{1}{n}\sum_{i=1}^{20}x_{i}   $$ \displaystyle 10=\frac{1}{20}\sum_{i=1}^{20}x_{i}   $$ \displaystyle \Rightarrow \sum_{i=1}^{20}x_{i}=200  $So, the  incorrect sum of observations $= 200$Correct sum of observation $= 200 - 8 = 192$$ \displaystyle \Rightarrow$ Correct mean $=\cfrac{Correct\, sum}{19}=\cfrac{192}{19}= 10.1  $Standard deviation $ \displaystyle \sigma =\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\frac{1}{n^{2}}\left ( \sum_{i=1}^{n}x_{i} \right )^{2}} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\left ( \bar{x} \right )^{2}}  $$ \displaystyle \Rightarrow 2=\sqrt{\frac{1}{20}Incorrrect \sum_{i=1}^{n}x_{i}^{2}-(10)^{2}}  $$ \displaystyle \Rightarrow 4=\frac{1}{20}Incorrect\, \sum_{i=1}^{n}x_{i}^{2}-100  $$ \displaystyle \Rightarrow Incorrect\sum_{i=1}^{n}x_{i}^{2}=2080  $$ \displaystyle \therefore  $ Correct $ \displaystyle \sum_{i=1}^{n}x_{i}^{2}=Incorrect \sum_{i=1}^{n}x_{i}^{2}-\left ( 8 \right )^{2}  $$= 2080 - 64$$= 2016$$ \displaystyle \therefore$ Correct Standard deviation $=\sqrt{\frac{Correct\sum x_{i}^{2}} {n}-(Correct\, Mean)^{2}}   $=$ \displaystyle \sqrt{\frac{2016}{19}-(10.1)^{2}}   $$ \displaystyle \sqrt{106.1-102.01}   $= $\displaystyle \sqrt{4.09}$= 2.02(ii) When $8$ is replaced by $12$Incorrect sum of observation $= 200$$ \displaystyle \therefore    $ Correct sum of observations $= 200 - 8 + 12 = 204$$ \displaystyle \therefore $ Correct mean = $ \displaystyle \frac{Correct sum}{20}=\frac{204}{20}= 10.2    $Standard deviation $ \displaystyle \sigma = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\frac{1}{n^{2}}\left ( \sum_{i=1}^{n}x_{i} \right )^{2}}=\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}- \left (\bar{x}  \right )^{2} }   $$ \displaystyle \Rightarrow $$ \displaystyle  2=\sqrt{\frac{1}{20}Incorrect \sum_{i=1}^{n}x_{i}^{2}-(10)^{2}}    $$ \displaystyle \Rightarrow $ $ \displaystyle 4 =\frac{1}{20}Incorrect \sum_{i=1}^{n}x_{i}^{2}-100   $$ \displaystyle \Rightarrow Incorrect\sum_{i=1}^{n}x_{i}^{2}=2080   $$ \displaystyle \therefore Correct\, \sum_{i=1}^{n}x_{i}^{2}=Incorrect \sum_{i=1}^{n}x_{i}^{2}-(8)^{2}+(12)^{2}   $$= 2080 - 64 + 144$$= 2160$$ \displaystyle Correct\, standard\, deviation\, =\sqrt{\frac{Correct\sum x_{i}^{2}}{n}-(Correct\, mean)^{2}}    $= $ \displaystyle \sqrt{\frac{2160}{20}-(10.2)^{2}}    $$ \displaystyle =\sqrt{108-104.04}    $$ \displaystyle =\sqrt{3.96}    $$= 1.98$

Question 4

The standard deviation of $1, 2, 3, 4, 5, 6, 7$ is?

Accepted Answer

Given observation is $1,2,3,4,6,7$Mean=$\cfrac { 1+2+3+4+5+6+7 }{ 7 } =\cfrac { 28 }{ 7 } =4$Variance=$\sum _{ +r=1 }^{ n }{ \cfrac { \left( x-{ x }_{ i } \right) ^{ 2 } }{ n }  } =\cfrac { \left( 1-4 \right) ^{ 2 }+\left( 2-4 \right) ^{ 2 }+\left( 3-4 \right) ^{ 2 }+\left( 4-4 \right) ^{ 2 }+\left( 5-4 \right) ^{ 2 }+\left( 6-4 \right) ^{ 2 }+\left( 7-4 \right) ^{ 2 } }{ 7 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { { 3 }^{ 2 }+{ 2 }^{ 2 }+{ 1 }^{ 2 }+0+{ 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 7 } \ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\cfrac { 9+4+1+1+4+9 }{ 7 } =\cfrac { 28 }{ 7 } =4$standard deviation=$\sqrt { variance } \ \sqrt { 4 } =2$

Question 5

Which of the following is not the measure of dispersion.

Accepted Answer

Measure of dispersion is the extent to which the distribution is stretched or squeezed.The common measures of dispersion are standard deviation, variance, interquartile range , mean deviation.

Question 6

The sum of the deviations of the variates  from the arithmetic mean is always.

Accepted Answer

The sum of the deviations of a given set of observations from their arithmetic mean is always zero. It is due to the property that the arithmetic mean is characterised as the centre of gravity. i.e. sum of positive deviation from the mean is equal to the sum of negative deviations.For example:$3,4,6,8,14$$\overline{x}=\dfrac{3+4+6+8+14}{5}=7$$x_{i}$       $x_{i}-\overline{x}$$3$          $-4$$4$          $-3$$6$          $-1$$8$             $1$$14$           $7$ $\sum (x_{i} - \overline{x})=-8+8=0$Hence, the sum of the deviations about mean is $0$.

Question 7

If the mean of the observations:$x, x + 3, x + 5, x + 7, x + 10$ is $9$, the mean of the last three observations is

Accepted Answer

Mean of $x, x + 3, x + 5, x + 7, x + 10 = 9$Sum of the terms $= 5x + 25$Mean = $\dfrac{5x+25}{5} =9$$x +5 =9$$x =4$Last three observations are $9,11,14$Mean $= \dfrac{9+11+14}{3}= 11\dfrac{1}{3}$

Question 8

The range of the data 25,18,20,22,16,6,17,12,30,32,10,19,8,11,20 is

Accepted Answer

The range of the data=Highest vale-lowest valueHighest value= 36Lowest value=6$\therefore$Range of the data=$32-6=24$

Question 9

Find the mean and variance for the following frequency distrubutionClasses0-1010-2020-3030-4040-50Frequencies5815166

Accepted Answer

Class interval$ f_i$$x_i$ $u_i=\cfrac{x_i-A}{h}$ $u_i^2$ $f_iu_i$  $f_iu_i^2$ 0-10 5 5 -2 4 -10 20 10-20 8 15 -1 1 -8 8 20-30 1525  0 0 0 0 30-40 1635  1 1 16 16 40-50 6 452  4 12 24 Total  50    10 68We have $N=\sum f_i=50$ Width of class $h= 10$Assumed mean  $A=25 (say)$Mean, $ \displaystyle \bar{x}=A+\frac{\sum_{i=1}^{5}f_{i}u_{i}}{N}\times h$$=25+\cfrac{10}{50}\times 10=25+2=27 $$\Rightarrow \bar{x}=27$Variance $\sigma ^{ 2 }=h^{ 2 }\left[ \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i }^{ 2 } }{ N } -\left( \cfrac { \sum _{ i=1 }^{ 5 } f_{ i }u_{ i } }{ N }  \right) ^{ 2 } \right] $$\sigma ^{ 2 }=10^{ 2 }\left[ \cfrac { 68 }{ 50 } -\left( \cfrac { 10 }{ 50 }  \right) ^{ 2 } \right] $$=10^{ 2 }\cfrac { \left[ 50\times 68-(10)^{ 2 } \right]  }{ \left( 50 \right) ^{ 2 } } $$ =\displaystyle \frac{1}{25}\left [ 3400-100 \right ]$$=\cfrac{3300}{25} $$=132$

Question 10

Which type of average is most affected by extreme values in the data?

Accepted Answer

Arithmetic mean refers to the average amount in a given group of data. It is defined as the summation of all the observation in the data which is divided by the number of observations in the data. Therefore, mean is affected by the extreme values because it includes all the data in a series.