Statistics

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

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

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

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

2. Mean of the given observation is.

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

$=\frac{500}{10}=50.$

So,

Therefore, the required mean deviation about the mean is

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

$=\frac{12+20+2+10+8+5+13+4+4+6}{10}$

$=\frac{84}{10}$

= 8.4

1. Mean of the given observation is.

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

$=\frac{80}{8}=10.$

Deviation of the respective observation about the mean $\overline{x}$ i.e.,  $x_i-\overline{x}$ are 4–10,7–10,8–10,9–10,10–10,12–10,13–10,17–10

=6, -3, -2, -1,0,2,3,7

The absolute value of the deviation i.e.,  $\left|x_i-\overline{x}\right|$ are 6,3,2,1,0,2,3,7.

Therefore, the required mean deviation about the mean is

$\mathrm{M.D}=(\overline{x})=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|x_i-\overline{x}\right|}=\frac{6+3+2+1+0+2+3+7}{8}$

$=\frac{24}{8}$

= 3.

There are lots of option for a UGC NET  apsirant to choose different subjects for paper 2 in their exam but as the person is M.Sc in mathematics so UGC NET paper 2 which does not have maths as a option for that student/aspirant some of the subjects are recommended to opt for their paper 2 such subjects are 

1) Economics  2) Management in finanace/business studies  3) Computer Science and Application  4) Education 

these are the four most important subjects for the students who are preparing for UGC NET in paper 2 

Topics to Study in IIT JAM Mathematical Statistics:

Aspirants can go through important topics that they should study to secure good marks in the IIT JAM exam for Mathematical Statistics below: