16, 38.68, 6.22

60.25, 27.4, 5.23

2.87

\(\frac { 52 }{ 7 } \)

Consider first n natural number, when n is even i.e. 1,2,3,4, ..., n (even).
Clearly, Mean \(\left( \bar { x } \right) =\frac { 1+2+3+...+n }{ n } =\frac { n\left( n+1 \right) }{ 2n } =\frac { n+1 }{ 2 } \)
\(MD=\frac { 1 }{ n } \left[ \left| 1-\frac { n+1 }{ 2 } \right| +\left| 2-\frac { n+1 }{ 2 } \right| +\left| 3-\frac { n+1 }{ 2 } \right| +......+\left| \frac { n-2 }{ 2 } -\frac { n+1 }{ 2 } \right| +\left| \frac { n }{ 2 } -\frac { n+1 }{ 2 } \right| +\left| \frac { n+2 }{ 2 } -\frac { n+1 }{ 2 } \right| +...\left| n-\frac { n+1 }{ 2 } \right| \right] \)
\(=\frac { 1 }{ n } \left[ \left| \frac { 1-n }{ 2 } \right| +\left| \frac { 3-n }{ 2 } \right| +\left| \frac { 5-n }{ 2 } \right| +...+\left| \frac { -3 }{ 2 } \right| +\left| \frac { 1 }{ 2 } \right| +...+\left| \frac { n-1 }{ 2 } \right| \right] \)
\(=\frac { 2 }{ n } \left[ \underbrace { \frac { 1 }{ 2 } +\frac { 3 }{ 2 } +...+\frac { n-1 }{ 2 } }_{ \left( \frac { n }{ 2 } \right) terms } \right]\)
\( =\frac { 1 }{ n } .{ \left( \frac { n }{ 2 } \right) }^{ 2 }\quad \left[ \because \ sum\ of\ first\ N\ odd\ natural\ numbers={ N }^{ 2 } \right]\)
\(=\frac { 1 }{ n } .\frac { { n }^{ 2 } }{ 4 } =\frac { n }{ 4 } \)

Let the assumed mean be a=64
Now,let as make a table from the given data.

Here, N=\(N=\sum { { f }_{ i } } =100;\sum { { f }_{ i } } { d }_{ i }=0\quad and\quad \sum { { f }_{ i }{ d }_{ i }^{ 2 } } =286\)
\(Now,variance({ \sigma ) }^{ 2 }=\left[ \frac { 1 }{ N } \sum { { f }_{ i }{ d }_{ i }^{ 2 } } -\left( \frac { 1 }{ N } \sum { { f }_{ i } } { d }_{ i } \right) ^{ 2 } \right] \)
\(\\ =\frac { 1 }{ 100 } \times 286-\left( \frac { 1 }{ 100 } \times 0 \right) ^{ 2 }=2.86\)
\(and\ standard\ deviation,\sigma =\sqrt { 2.86 } =1.69\)

The given data can be arranged in ascending order as 30,34,38,40,42,44,50,51,60,66.
Here, total number of observations are 10 i.e. n = 10, which is even
\(\therefore \) Median
\(M=\frac { \left( \frac { n }{ 2 } \right) th\quad observation\quad +\left( \frac { n }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { \left( \frac { 10 }{ 2 } \right) th\quad observation\quad +\left( \frac { 10 }{ 2 } +1 \right) th\quad observation\quad }{ 2 } \)
\(=\frac { 5th\quad observation\quad +6th\quad observation\quad }{ 2 } \)
\(=\frac { 42+44 }{ 2 } =\frac { 86 }{ 2 } =43\)
Let us make the table for absolute deviation

Now, mean deviation about the median.
\(MD=\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-M \right| } }{ 10 } =\frac { 87 }{ 10 } =8.7\)

Given observations are
6.5,5,5.25,5.5,4.75,4.5,6.25,7.75,8.5
Here number of observations, n = 9
Let \(\overline { x } \) be the mean of given data.
Then, \(\overline { x } =\frac { \left( 6.5+5+5.25+5.5+4.75+4.5+6.25+7.75+8.5 \right) }{ 9 } =\frac { 54 }{ 9 } =6\)
Let us make the table for deviation and absolute deviation

\(\therefore \)  Mean deviation about mean,
\(MD\left( \overline { x } \right) =\frac { \sum _{ i-1 }^{ 10 }{ \left| { x }_{ i }-\overline { x } \right| } }{ 9 } =\frac { 10 }{ 9 } =1.1\)
Hence, the mean deviation about mean is 1.1

We have known that, if any constant is added in each observation, then standard deviation remains same.
So, the standard deviation of the observations a + k,d + k,c + k,d + k,e + k is S

Take the assumed mean a =45 and h = 10 Make the table for step deviation and product of frequency with absolute deviation

 Here, \(\sum _{ i=1 }^{ 7 }{ { f }_{ i }=40 } ,\sum _{ i=1 }^{ 7 }{ { f }_{ i }{ u }_{ i }=0 } and\quad \sum { { f }_{ i }|{ x }_{ i }-\bar { x } | } =400\)
\(Now,\quad \bar { x } =a+\frac { \sum _{ i=1 }^{ 7 }{ { f }_{ i }{ u }_{ i } } }{ \sum _{ i=1 }^{ 7 }{ { f }_{ i } } } \times h=45+\frac { 0 }{ 40 } \times 10=45\)
\(\therefore \) Mean deviation about mean
\(=\frac { 1 }{ \sum _{ i=1 }^{ 7 }{ { f }_{ i } } } \sum _{ i=1 }^{ 7 }{ { f }_{ i }|{ x }_{ i }-\bar { x } |=\frac { 400 }{ 40 } =10 } \)

Let us make the following table from the given data.

\(\therefore \ (\bar { x } )=\frac { \sum { { f }_{ i }{ x }_{ i } } }{ N } =\frac { 760 }{ 40 } =19\)
\(and\ standard\ deviation=\sqrt { \frac { 1 }{ N } [\sum { { f }_{ i }({ x }_{ i }-\bar { x } )^{ 2 }] } } \)
\(=\sqrt { \frac { 1736 }{ 40 } } =\sqrt { 434 } =6.59\)
Ans.19,6.59

Find the median and then calculate the mean deviation about the median by using the formula \(\frac { \sum { f_{ i }|x_{ i }-M| } }{ \sum { f_{ i } } } \)
Here, N = 30, which is even.
So, median is the mean of the 15th and 16th observations. Both of these observations lie in the cumulative frequency 18 for which the corresponding observations is 13.
Therefore, 
\(Median(M)=\frac { 15th\quad observation\quad +\quad 16th\quad observation }{ 2 }\)
\(=\frac { 13+13 }{ 2 } =13\)
We make the table from the given data.

We have, \(\sum { f_{ i }=30 } \) and \(\sum { f_{ i }|x_{ i }-M|=149 } \) 
\(\therefore \) Mean deviation about median,
\(MD(M)=\frac { 1 }{ N } \sum { f_{ i }| } x_{ i }-M|=\frac { 1 }{ 30 } \times 149=4.97\)

Here \(\sum _{ i=1 }^{ 50 }{ { x }_{ i }=212,\sum _{ i=1 }^{ 50 }{ { x }_{ i }^{ 2 } } =902.8, } \) \(\sum _{ i=1 }^{ 50 }{ { y }_{ i } } =261,\sum _{ i=1 }^{ 50 }{ { y }_{ i }^{ 2 } } =1457.6\)
Now \(\bar { x } =\frac { 212 }{ 50 } =4.24\)
\({ \sigma }_{ x }^{ 2 }=\frac { 1 }{ 50 } \times \times 902.8-{ \left( \frac { 212 }{ 50 } \right) }^{ 2 }\)
= 18.056 - 17.978 = 0.078
\({ \sigma }_{ x }=\sqrt { 0.078 } =0.28\)
Also \(\bar { y } =\frac { 261 }{ 50 } =5.22\)
\({ \sigma }_{ y }^{ 2 }=\frac { 1 }{ 50 } \times 1457.6-{ \left( \frac { 261 }{ 50 } \right) }^{ 2 }\)
= 29.152-27.248 = 1.904
\({ \sigma }_{ y }=\sqrt { 1.904 } =1.38\)
C.V. of length=\(\frac { 0.28 }{ 4.24 } \times 100=6.6\)
C.V. of weight=\(\frac { 1.38 }{ 5.22 } \times 100=26.45\)
C.V. of weight > C.V. of length
Thus weight have more variability than length.

Mean \((\overline { x } )=\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 22 } \times 2200=100\)
Variance = \({ \sigma }^{ 2 }=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }{ \left( { x }_{ i }-\overline { x } \right) }^{ 2 } } \)
\(=\frac { 1 }{ 22 } \times 640=29.09\)

Mean\(\overline { x } =\frac { 1 }{ N } \sum { { f }_{ i }{ x }_{ i } } =\frac { 1 }{ 50 } \times 17900=358\)
Mean deviation about mean\(=\frac { 1 }{ N } \sum _{ i=1 }^{ n }{ { f }_{ i }\left| { x }_{ i }-\overline { x } \right| } \)
\(=\frac{1}{50}\times7896 = 157.92\)