# Statistics

## Class 11 NCERT

### NCERT

1   Find the mean deviation about the mean for the data $4, 7, 8, 9, 10, 12, 13, 17$

##### Solution :

The given data is $4, 7, 8, 9, 10, 12, 13, 17$$\\ Mean of the data,\bar{x}=\dfrac{4+7+8+9+10+12+13+17}{8}\\ =\dfrac{80}{8}=10$$\\$ The deviations of the respective observations from the mean $\bar{x}$, i. e.,$x_i-\bar{x},$ are $-6,-3,-2,-1,0,2,3,7$$\\ The absolute values of the deviations, i.e. |x _i - \bar{x}|, , are6,3,2,1,0,2,3,7$$\\$ The required mean deviation about the mean is$\\$ $M.D.(\bar{x})=\dfrac{\displaystyle\sum_{i=1}^{8}|x_i-\bar{x}|}{8}\\ =\dfrac{6+3+2+1+0+2+3+7}{8}=\dfrac{24}{8}=3$

2   Find the mean deviation about the mean for the data $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$

The given data is $38, 70, 48, 40, 42, 55, 63, 46, 54, 44$ Mean of the given data,$\\$ $\bar{x}=\dfrac{38+70+48+40+55+63+46+54+44}{10}\\ =\dfrac{500}{10}=50$$\\ The deviations of the respective observations from the mean \bar{x}, i.e.,x_i-\bar{x}, are-12,20,-2,-10,-8,5,13,-4,4,-6$$\\$ The absolute values of the deviations,i.e.,$|x_i-\bar{x}|,$ are$12,20,2,10,8,5,13,4,4,6$$\\ The required mean deviation about the mean is \\ M.D.(\bar{x})=\dfrac{\displaystyle\sum_{i=1}^{8}|x_i-\bar{x}|}{10}\\ =\dfrac{12+20+2+10+8+5+13+4+4+6}{10}\\ =\dfrac{84}{10}=8.4 3 Find the mean deviation about the median for the data. 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17 ##### Solution : The given data is 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17$$\\$ Here, the numbers of observations are $12,$ which is even.$\\$ Arranging the data in ascending order, we obtain$\\$ $10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18$$\\ Median,M=\\ \dfrac{(\dfrac{12}{2})^{th}\text{observation}+(\dfrac{12}{2}+1)^{th}\text{observation}}{2}\\ =\dfrac{6^{th}\text{observation} + 7^{th}\text{observation} }{2}\\ =\dfrac{13+14}{2}\\ =\dfrac{27}{2}=13.5$$\\$ The deviations of the respective observations from the median,i.e.,$x_i-M,$ are$-3.5,-2.5,-2.5,-1.5,-0.5,-0.5,0.5,2.5,2.5,3.5,3.5,4.5$$\\ The absolute values of the deviations,|x_i-M| are \\ 3.5,2.5,2.5,1.5,0.5,0.5,0.5,2.5,2.5,3.5,3.5,4.5$$\\$ The required mean deviation about the median is $\\$ $M.D.(M)=\dfrac{\displaystyle\sum_{i=1}^{10}|x_i-M|}{12}\\ =\dfrac{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5}{12}\\ =\dfrac{28}{12}=2.33$

4   Find the mean deviation about the median for the data $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$

The given data is $36, 72, 46, 42, 60, 45, 53, 46, 51, 49$$\\ Here, the number of observations is 10, which is even. Arranging the data in ascending order, we obtain 36, 42, 45, 46, 46, 49, 51, 53, 60, 72$$\\$ Median $M=\\ \dfrac{(\dfrac{10}{2})^{th}\text{observation} +(\dfrac{10}{2}+1)^{th}\text{observation}}{2}\\ =\dfrac{5^{th}\text{observation} + 6^{th} \text{observation}}{2}\\ =\dfrac{46+49}{2}\\ =\dfrac{95}{2}=47.5$$\\ The deviations of the respective observations from the median, i.e. x_i-M are -11.5,-5.5,-2.5,-1.5,-1.5,1.5,3.5,5.5,12.5,24.5$$\\$ The absolute values of the deviations, $|x_i-M|,$ are$\\$ $11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5$$\\ Thus, the required mean deviation about the median is\\ M.D.(M)=\dfrac{ \displaystyle\sum_{i=1}^{10}|x_i-M|}{10}\\ =\dfrac{11.5,5.5,2.5,1.5,1.5,1.5,3.5,5.5,12.5,24.5}{10}\\ =\dfrac{70}{10}=7 5 Find the mean deviation about the mean for the data.\\ \begin{array}{|c|c|c|c|c|} \hline x_i&5&10&15&20&25 \\ \hline f_i&7&4&6&3&5 \\ \hline \end{array} ##### Solution : \begin{array}{|c|c|c|c|c|} \hline x_i & f_i & f_ix_i & |x_i-\bar{x}| & f_i|x_i-\bar{x}| \\ \hline 5 & 7 & 35 & 9 & 63 \\ \hline 10 & 4 & 40 & 4 & 16 \\ \hline 15 & 6 & 90 & 1 & 6 \\ \hline 20 & 3 & 60 & 6 & 18 \\ \hline 25 & 5 & 125 & 11 & 55 \\ \hline & 25 & 350 & & 158\\ \hline \end{array}$$\\$ $N = \displaystyle\sum_{i = 1}^{5} f_i = 25 \\ \displaystyle\sum_{i = 1}^{5} f_i x_i = 350 \\ \therefore \bar{x} = \dfrac{1}{N}\displaystyle\sum_{i = 1}^{5}f_i x_i \\ = \dfrac{1}{25} * 350 = 14 \\ \therefore MD(\bar{x}) = \dfrac{1}{N} \displaystyle\sum_{i = 1}^{5} f_i |x_i - \bar{x}| \\ \dfrac{1}{25} * 158 = 6.32$

6   Find the mean deviation about the mean for the data$\\$ $\begin{array}{|c|c|c|c|c|c|} \hline x_i & 10 & 30 & 50 & 70 & 90 \\ \hline f_i & 4 & 24 & 28 & 16 & 8 \\ \hline \end{array}$

##### Solution :

$\begin{array}{|c|c|c|c|c|c|} \hline x_i & f_i & f_i x_i & |x_i-\bar{x}|& f_i|x_i-\bar{x}| \\ \hline 10& 4 & 40 & 40& 160 \\ \hline 30 & 24 & 720 & 20 & 480 \\ \hline 50 & 28 & 1400 & 0 & 0 \\ \hline 70 & 16 & 1120 & 20 & 320 \\ \hline 90 & 8 & 720 & 40 & 320 \\ \hline & 80 & 4000 & & 1280 \\ \hline \end{array}$ $N=\displaystyle\sum_{i=1}^{5} f_i =80,\displaystyle\sum_{i=1}^{5} f_i x_i=4000\\ \therefore \bar{x}=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i x_i=\dfrac{1}{80}*4000=50\\ \therefore MD(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i |x_i-\bar{x}|=\dfrac{1}{80}*1280=16$

7   Find the mean deviation about the median for the data.$\\$ $\begin{array}{|c|c|c|c|c|c|c|} \hline x_i & 5 & 7 & 9 & 10 & 12 & 15 \\ \hline f_i & 8 & 6 & 2 & 2 & 2 & 6\\ \hline \end{array}$

##### Solution :

The given observations are already in ascending order.$\\$ Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.$\\$ $\begin{array}{|c|c|c|} \hline x_i & f_i & c.f. \\ \hline 5 & 8 & 8 \\ \hline 7 & 6 & 14 \\ \hline 9 & 2 & 16 \\ \hline 10 & 2 & 18 \\ \hline 12 & 2 & 18 \\ \hline 15 & 6 & 26 \\ \hline \end{array}$ Here, $N = 26$, which is even.$\\$ Median is the mean of $13^{th}$ and $14^{th}$ observations. Both of these observations lie in the cumulative frequency $14$, for which the corresponding observation is $7.$$\\ \therefore \text{Median } =\dfrac{13^{th}\text{observation} + 14^{th} \text{oservation}}{2}\\ =\dfrac{7+7}{2}=7$$\\$ The absolute values of the deviations from median, i.e.$|x _i - M |$, are $\\$ $\begin{array}{|c|c|c|c|c|c|c|} \hline |x_i-M|& 2 & 0 & 2 & 3 & 5 & 8\\ \hline f_i & 8 & 6 & 2 & 2 & 2 & 6 \\ \hline f_i|x_i-M| & 16 & 0 & 4 & 6 & 10 & 48 \\ \hline \end{array}$ $\displaystyle\sum_{i=1}^{6} f_i =26 \text{and} \displaystyle\sum_{i=1}^{6} f_i|x_i-M|=84\\ M.D.(M)=\dfrac{1}{N}\displaystyle\sum_{i=1}^{6} f_i|x_i-M|=\dfrac{1}{26}* 84 =3.23$

8   Find the mean deviation about the median for the data$\\$ $\begin{array}{|c|c|c|c|} \hline x_i & 15 & 21 & 27 & 30 & 35 \\ \hline f_i & 3 & 5 & 6 & 7 & 8 \\ \hline \end{array}$

##### Solution :

The given observations are already in ascending order.$\\$ Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.$\\$ $\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & c.f. \\ \hline 15 & 3 & 3 \\ \hline 21 & 5 & 8\\ \hline 27 & 6 & 14 \\ \hline 30 & 7 & 21\\ \hline 35 & 8 & 29 \\ \hline \end{array}$ $\\$ Here,$N=29 ,$ which is odd $\\$ $\therefore$ Median$=(\dfrac{7+7}{2})^{th}$ observation $=15 ^{th}$ observation $\\$ This observation lies in the cumulative frequency $21$, for which the corresponding observation is $30.$$\\ \therefore \text{Median} =30$$\\$ The absolute values of the deviations from median, i.e. $|x_ i -M|$ , are $\\$ $\begin{array}{|c|c|c|c|c|} \hline |x_i-M| & 15 & 9 & 3 & 0 & 5\\ \hline f_i & 3 & 5 & 6 & 7 & 8 \\ \hline f_i|x_i-M| & 45 & 45 & 18 & 0 & 40 \\ \hline \end{array}$$\\$ $\displaystyle\sum_{i=1}^{5} f_i =29, \displaystyle\sum_{i=1}^{5} f_i |x_i-M|=148\\ \therefore M.D.(M)=\dfrac{1}{N}\displaystyle\sum_{i=1}^{5} f_i |x_i-M|=\dfrac{1}{29}*148=5.1$

9   Find the mean deviation about the mean for the data.$\\$ $\begin{array}{|c|c|c|c|c|} \hline \text{Income per day} & \text{Number of persons} \\ \hline 0-100 & 4 \\ \hline 100 - 200 & 8 \\ \hline 200 - 300 & 9 \\ \hline 300 -400 & 10 \\ \hline 400 -500 & 7 \\ \hline 500 -600 & 5 \\ \hline 600-700 & 4 \\ \hline 700 -800 &3 \\ \hline \end{array}$

##### Solution :

The following table is formed.$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Income per day} & \text{Number of persons f_i} & Mid-point x_i & f_ix_i & |x_i-\bar{x}| & f_i|x_i-\bar{x}| \\ \hline 0-100 & 4 & 50 & 200 & 308 & 1232 \\ \hline 100-200 & 8 & 150 & 1200 & 208 & 1664 \\ \hline 200 -300 & 9 & 250 & 2250 & 108 7 972 \\ \hline 300-400 & 10 & 350 & 3500 & 8 & 80 \\ \hline 400 -500 & 7 & 450 & 3150 & 92 & 644 \\ \hline 500-600 & 5 & 550 & 2750 & 192 & 960 \\ \hline 600-700 & 4 & 650 & 2600 & 292 & 1168 \\ \hline 700 -800 & 3 7 750 & 2250 & 392 & 1176 \\ \hline & 50 & 17900 & & 7896 \\ \hline \end{array}$ $\\$ Here, $N=\displaystyle\sum_{i=1}^{8} f_i =50,\displaystyle\sum_{i=1}^{8} f_ix_i=17900\\ \therefore \bar{x}=\dfrac{1}{N}\displaystyle\sum_{i=1}^{8} f_ix_i=\dfrac{1}{50}*17900=358\\ M.D.(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{8} f_i|x_i-\bar{x}|=\dfrac{1}{50}*7896=157.92$

10   Find the mean deviation about the mean for the data$\\$ $\begin{array}{|c|c|c|c|} \hline \text{Height in cms} & \text{Number of boys}\\ \hline 95-105 & 9 \\ \hline 105-115 & 13 \\ \hline 115-125 & 26 \\ \hline 125- 135 & 30 \\ \hline 135 - 145 & 12 \\ \hline 145 - 155 & 10 \\ \hline \end{array}$

##### Solution :

The following table is formed.$\\$ $\begin{array}{|c|c|c|c|c|c|} \hline \text{Height in cms} & \text{Number of boys f_i} & \text{Mid-point}x_i & f_i x_i & |x_i-\bar{x}|& f_i|x_i-\bar{x}|\\ \hline 95-105 & 9 & 100 & 900 & 25.3 & 227.7\\ \hline 105-115 &13 &110 &1430 & 15.3 &198.9 \\ \hline 115-125 & 26 &120 & 3120 & 5.3 & 137.8 \\ \hline 125-135 & 30 & 130 & 3900 & 4.7 & 141\\ \hline 135-145 & 12 & 140 & 1680 & 14.7 & 176.4 \\ \hline 145-155 7 10 & 150 & 1500 & 24.7 & 247\\ \hline \end{array}$ $\\$ Here,$N=\displaystyle\sum_{i=1}^{6} f_i =100, \displaystyle\sum_{i=1}^{6} f_ix_i=12530\\ \therefore \bar{x} =\dfrac{1}{N} \displaystyle\sum_{i=1}^{6} f_ix_i=\dfrac{1}{100}* 12530=125.3\\ M.D.(\bar{x})=\dfrac{1}{N}\displaystyle\sum_{i=1}^{6} f_i|x_i-\bar{x}|=\dfrac{1}{100}*1128.8= 11.28$

11   Calculate the mean deviation about median age for the age distribution of 100 persons given below: $\\$ $\begin{array}{|c|c|c|c|} \hline \text{Age} & \text{Number }\\ \hline 16-20& 5 \\ \hline 21-25 & 6 \\ \hline 26-30 & 12 \\ \hline 31-35 &14 \\ \hline 36-40 &26 \\ \hline 41-45 & 12\\ \hline 46-50 & 16\\ \hline 51-55&9\\ \hline \end{array}$ $\begin {array}{|c|c|c|c|} \hline \text{Age} & \text{Number }f_i& \text{Cumulative frequency(c.f)}& \text{Mid-point }x_i& |x_i-\text{Med}|& fi|x_i-\text{Med}\\ \hline 15.5-20.5& 5 & 5& 18&20& 100\ \hline 20.5-25.5 & 6& 11 & 23& 15&99\\ \hline 25.5-30.5 & 12& 23 & 28&10& 120\\ \hline 30.5-35.5&14 & 37& 33& 5& 70\\ \hline 35.5-40.5 &26& 63&38& 0& 0 \\ \hline 40.5-45.5 & 12& 75& 43& 5&60\\ \hline 45.5-50.5 & 16& 91& 48& 10&160\ \hline 50.5-55.5&9& 100& 53& 15&135\\ \hline & 10&&&&735\\ \hline \end{array}$

##### Solution :

The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.$\\$ The table is formed as follows.$\\$\\ $\begin {array}{|c|c|c|c|}\\ \hline \text{Age} & \text{Number }f_i& \text{Cumulative frequency(c.f)}& \text{Mid-point }x_i& |x_i-\text{Med}|& fi|x_i-\text{Med}\\ \hline 15.5-20.5& 5 & 5& 18&20& 100\ \hline 20.5-25.5 & 6& 11 & 23& 15&99\\ \hline 25.5-30.5 & 12& 23 & 28&10& 120\\ \hline 30.5-35.5&14 & 37& 33& 5& 70\\ \hline 35.5-40.5 &26& 63&38& 0& 0 \\ \hline 40.5-45.5 & 12& 75& 43& 5&60\\ \hline 45.5-50.5 & 16& 91& 48& 10&160\ \hline 50.5-55.5&9& 100& 53& 15&135\\ \hline & 10&&&&735\\ \hline \end{array}$