CBSE Class 11 Mathematics Chapter 15: Statistics NCERT Solutions

The given data set consists of 8 observations: 4, 7, 8, 9, 10, 12, 13, 17.

First, we calculate the mean ($\backslash overline\{x\}$) of the data:

$$\backslash overline\{x\}\; =\; \backslash frac\{4+7+8+9+10+12+13+17\}\{8\}\; =\; \backslash frac\{80\}\{8\}\; =\; 10$$

Next, we find the deviation of each observation from the mean ($x\_i\; -\; \backslash overline\{x\}$):

$-6,\; -3,\; -2,\; -1,\; 0,\; 2,\; 3,\; 7$

Now, we find the absolute values of these deviations ($|x\_i\; -\; \backslash overline\{x\}|$):

$6,\; 3,\; 2,\; 1,\; 0,\; 2,\; 3,\; 7$

Finally, we calculate the mean deviation about the mean (M.D.) by taking the average of these absolute deviations:

$$M.D.(\backslash overline\{x\})\; =\; \backslash frac\{\backslash sum\_\{i=1\}^\{8\}\; |x\_i\; -\; \backslash overline\{x\}|\}\{8\}\; =\; \backslash frac\{6+3+2+1+0+2+3+7\}\{8\}\; =\; \backslash frac\{24\}\{8\}\; =\; 3$$

The mean deviation about the mean is 3.

The given data set has 10 observations: 38, 70, 48, 40, 42, 55, 63, 46, 54, 44.

First, we compute the mean ($\backslash overline\{x\}$) of this data:

$$\backslash overline\{x\}\; =\; \backslash frac\{38\; +\; 70\; +\; 48\; +\; 40\; +\; 42\; +\; 55\; +\; 63\; +\; 46\; +\; 54\; +\; 44\}\{10\}\; =\; \backslash frac\{500\}\{10\}\; =\; 50$$

Next, we determine the deviations of each observation from the mean ($x\_i\; -\; \backslash overline\{x\}$):

$-12,\; 20,\; -2,\; -10,\; -8,\; 5,\; 13,\; -4,\; 4,\; -6$

Then, we find the absolute values of these deviations ($|x\_i\; -\; \backslash overline\{x\}|$):

$12,\; 20,\; 2,\; 10,\; 8,\; 5,\; 13,\; 4,\; 4,\; 6$

Finally, we calculate the mean deviation about the mean (M.D.) by averaging these absolute deviations:

$$M.D.(\backslash overline\{x\})\; =\; \backslash frac\{\backslash sum\_\{i=1\}^\{10\}\; |x\_i\; -\; \backslash overline\{x\}|\}\{10\}\; =\; \backslash frac\{12+20+2+10+8+5+13+4+4+6\}\{10\}\; =\; \backslash frac\{84\}\{10\}\; =\; 8.4$$

The mean deviation about the mean for this data is 8.4.

The given data set has 12 observations: 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17.

First, we arrange the data in ascending order:

10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

Since there are 12 observations (an even number), the median (M) is the average of the $\backslash frac\{12\}\{2\}$ = 6th and $\backslash left(\backslash frac\{12\}\{2\}\; +\; 1\backslash right)$ = 7th observations.

$$M\; =\; \backslash frac\{6^\{th\}\; \backslash text\{\; observation\}\; +\; 7^\{th\}\; \backslash text\{\; observation\}\}\{2\}\; =\; \backslash frac\{13\; +\; 14\}\{2\}\; =\; \backslash frac\{27\}\{2\}\; =\; 13.5$$

Next, we find the deviation of each observation from the median ($x\_i\; -\; M$):

$-3.5,\; -2.5,\; -2.5,\; -1.5,\; -0.5,\; -0.5,\; 0.5,\; 2.5,\; 2.5,\; 3.5,\; 3.5,\; 4.5$

Now, we find the absolute values of these deviations ($|x\_i\; -\; M|$):

$3.5,\; 2.5,\; 2.5,\; 1.5,\; 0.5,\; 0.5,\; 0.5,\; 2.5,\; 2.5,\; 3.5,\; 3.5,\; 4.5$

Finally, we calculate the mean deviation about the median (M.D.) by taking the average of these absolute deviations:

$$M.D.(M)\; =\; \backslash frac\{\backslash sum\_\{i=1\}^\{12\}\; |x\_i\; -\; M|\}\{12\}\; =\; \backslash frac\{3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5\}\{12\}$$

$$M.D.(M)\; =\; \backslash frac\{28\}\{12\}\; =\; \backslash frac\{7\}\{3\}\; \backslash approx\; 2.33$$

The mean deviation about the median is approximately 2.33.