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 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} ight )^{2}} = \sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}-\left ( \bar{x} ight )^{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 	herefore    Correct   \displaystyle \sum_{i=1}^{n}x_{i}^{2}=Incorrect \sum_{i=1}^{n}x_{i}^{2}-\left ( 8 ight )^{2}   = 2080 - 64 = 2016  \displaystyle 	herefore  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 	herefore      Correct sum of observations  = 200 - 8 + 12 = 204  \displaystyle 	herefore   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} ight )^{2}}=\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_{i}^{2}- \left (\bar{x}  ight )^{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 	herefore 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 3

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 } ight) ^{ 2 } }{ n }  } =\cfrac { \left( 1-4 ight) ^{ 2 }+\left( 2-4 ight) ^{ 2 }+\left( 3-4 ight) ^{ 2 }+\left( 4-4 ight) ^{ 2 }+\left( 5-4 ight) ^{ 2 }+\left( 6-4 ight) ^{ 2 }+\left( 7-4 ight) ^{ 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 4

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 5

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 6

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 	herefore Range of the data= 32-6=24

Question 7

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} 	ext{ observations}}{2} 63 = \dfrac{x+x+2}{2} \Rightarrow x+1 = 63 \Rightarrow x=62

Question 8

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 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}	imes h =25+\cfrac{10}{50}	imes 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 }  ight) ^{ 2 } ight]  \sigma ^{ 2 }=10^{ 2 }\left[ \cfrac { 68 }{ 50 } -\left( \cfrac { 10 }{ 50 }  ight) ^{ 2 } ight]  =10^{ 2 }\cfrac { \left[ 50	imes 68-(10)^{ 2 } ight]  }{ \left( 50 ight) ^{ 2 } }   =\displaystyle \frac{1}{25}\left [ 3400-100 ight ] =\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.