Let us make the following table from the given data.

Here, \(\frac { N }{ 2 } =50\) 
So, the median class is 35.5 - 40.5
\(\therefore \)  l = 35.5, cf = 37, f = 26 and h = 5
Now,
\(Median\left( M \right) =l+\frac { \frac { N }{ 2 } -cf }{ f } \times h\) 
\(=35.5+\frac { 50-37 }{ 26 } \times 5=35.5+\frac { 13 }{ 26 } \times 5=35.5+2.5=38\)   
and Mean deviation about median
\(=\frac { \Sigma fi|{ x }_{ i }-M| }{ \Sigma fi } =\frac { 735 }{ 100 } =7.35\)  
Hence, the mean deviation about median is 7.35.

On arranging the given data in ascending order, we have 22,24,25,27,28,29,30,31,41,42
Now, Mean
\(\bar { x } =\frac { 22+24+25+27+28+29+30+31+41+42 }{ 10 } =\frac { 299 }{ 10 } =29.9\)
and mean deviation from the mean, \(MD\left( \bar { x } \right) =\frac { \sum _{ i=1 }^{ n }{ \left| { { x }_{ i }-\bar { x } } \right| } }{ n } \)
\(=\frac { \left[ \left| 22-29.9 \right| +\left| 24-29.9 \right| +\left| 25-29.9 \right| +\left| 27-29.9 \right| +\left| 28-29.9 \right| +\left| 29-29.9 \right| +\left| 30-29.9 \right| +\left| 31-29.9 \right| +\left| 41-29.9 \right| +\left| 42-29.9 \right| \right] }{ 10 } \)
\(=\frac { 7.9+5.9+4.9+2.9+1.9+0.9+0.1+1.1+11.1+12.1 }{ 10 } \)
\(=\frac { 48.8 }{ 10 } =4.88=4.9\left( approx \right) .\)
\(\therefore \ \bar { x } -MD\left( \bar { x } \right) =29.9-4.9=25\)
\(\bar { x } +MD\left( \bar { x } \right) =29.9+4.9=34.8\)
On examining the arranged data, we find that observations between 25 and 34.8 are 27,28,29,30,31.
Hence, there are 5 observations lying between \(\bar { x } -MD\left( \bar { x } \right) \) and \(\bar { x } +MD\left( \bar { x } \right) \)

The first n natural numbers are 1,2,3...,n.
\(\because\) Standard deviation,
\(SD=\sqrt { \frac { \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } }{ n } -({ \frac { \sum _{ i=1 }^{ n }{ { x }_{ i } } }{ n } ) }^{ 2 } } \)
\(\therefore \ SD=\sqrt { \frac { (n(n+1)(2n+1) }{ 6n } -({ \frac { n(n+1) }{ 2n } ) }^{ 2 } } \)
\([\because \ \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 } and\ \sum _{ i=1 }^{ n }{ { x }_{ i } } =\frac { n(n+1) }{ 2 } ]\)
\(=\sqrt { (n+1)(\frac { 2n+1 }{ 6 } -\frac { n+1 }{ 4 } ) } \)
\(=\sqrt { (n+1)(\frac { 4n+2-3n-3 }{ 12 } ) } \)
\(=\sqrt { \frac { (n+1)(n-1) }{ 12 } } =\sqrt { \frac { { n }^{ 2 }-1 }{ 12 } } \)
\(\therefore \ Variance={ (SD) }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 12 } \)