Ncert Solutions Maths class 11th

12. The given data is made'continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. So, we cam tabulate as.

11. From the given data we can tabulate the following.

10. From the given data we can insulate the following.

Take the assumed mean a=120 and h=10

Heigts in cm.	No. of boys fi	Mid-pointsxi		fidi	|xi - x? |	fi |xi - x? |
95-105	9	100	-2	-18	25.3	227.7
105-115	13	110	-1	-13	15.3	198.9
115-125	26	120	0	5.3	137.8
125-135	30	130	1	30	4.7	141
135-145	12	140	2	24	14.7	176.4

9. From the given data we cantabulate the following.

8. From the given data we can tabulate the following.

Here N = 29 which is odd.

= 1/29 * 148

= 5. 10

7. From the given data we cantabulate the following.

Now, N=26 which is even.

So, Median is the mean of 13^th and 14^th observation. Both of these observations lie in the cumulative frequency 14 for which corresponding observation is 7.

6. From the given data we tabulate the following.

We have,

5. From the given data we have,

4. Arranging the given data in ascending order we get,

36,42,45,46,46,49,51,53,60,72

As n = 10 (even)

$=\frac{\left(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\text{\hspace{0.17em}thObservation}}^{}+{(\raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+1)thObservation}^{}}{2}$

$\Rightarrow =\frac{{\mathrm{5thobservation}}^{}+{\mathrm{6thobservation}}^{}}{2}$

$=\frac{46+49}{2}$

$=\frac{95}{2}$

= 47.5

$\therefore$ M.D. (M) $=\frac{1}{n}*{\displaystyle \sum _{i=1}^n\left|x_i-M\right|}$

$=\frac{1}{10}*\left(11.5+5.5+2.5+1.5+1.5+1.5+3.5+5.5+12.5+24.5\right)$

$=\frac{70}{10}=7.$

3. Arranging the data in ascending order we get,

10,11,1112,13,13,14,16,16,17,17,18

As n=12, even

So, median is the mean of ${(\frac{M}{2})th}^{}$ and ${(\frac{M}{2}+1)th}^{}$ observation.

$\Rightarrow =\frac{{\mathrm{6thobservation}}^{}+{\mathrm{7thobservation}}^{}}{2}$

$M=\frac{13+14}{2}=\frac{27}{2}=13.5.$

So, deviation of respective observation about the median. $M,\left|x_i-M\right|$ are

Therefore the mean deviation about the mean is

$\text{\hspace{0.17em}M.D.}(M)=\frac{1}{n}*{\displaystyle \sum _{i=1}^n\left|x_i-M\right|}$

$=\frac{1}{12}*(3.5+2.5+2.5+1.5+0.5+0.5+0.5+2.5+2.5+3.5+3.5+4.5)$

$=\frac{28}{12}=2.33.$