CBSE 11th Standard Maths Subject Statistics HOT Questions 2 Mark Questions With Solution 2021
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CBSE 11th Standard Maths Subject Statistics HOT Questions 2 Mark Questions With Solution 2021
11th Standard CBSE
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Reg.No. :
Mathematics
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Find the standard deviation and the variance of first n natural numbers.
(a) -
If \(\bar { x } \) is mean and \(MD\left( \bar { x } \right) \) is the mean deviation from mean then find the number of observations lying between \(\bar { x } -MD\left( \bar { x } \right) \) and \(\bar { x } +MD\left( \bar { x } \right) \) . Use the data 22,24,30,27,29,31,25,28,41,42.
(a) -
Calculate the mean deviation about median for the age distribution of 100 persons given below.
Age 16-20 21-25 26-30 31-35 36-40 41-45 46-50 51-55 Number 5 6 12 14 26 12 16 9 First, make the class intervals are of uniform length and then use the following formula.
\(Median\left( M \right) =l+\frac { \frac { N }{ 2 } -cf }{ f } \times h\quad and\quad MD\left( M \right) =\frac { \Sigma fi|{ x }_{ i }-M| }{ \Sigma fi } \)(a)
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CBSE 11th Standard Maths Subject Statistics HOT Questions 2 Mark Questions With Solution 2021 Answer Keys
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The first n natural numbers are 1,2,3...,n.
\(\because\) Standard deviation,
\(SD=\sqrt { \frac { \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } }{ n } -({ \frac { \sum _{ i=1 }^{ n }{ { x }_{ i } } }{ n } ) }^{ 2 } } \)
\(\therefore \ SD=\sqrt { \frac { (n(n+1)(2n+1) }{ 6n } -({ \frac { n(n+1) }{ 2n } ) }^{ 2 } } \)
\([\because \ \sum _{ i=1 }^{ n }{ { x }_{ i }^{ 2 } } =\frac { n(n+1)(2n+1) }{ 6 } and\ \sum _{ i=1 }^{ n }{ { x }_{ i } } =\frac { n(n+1) }{ 2 } ]\)
\(=\sqrt { (n+1)(\frac { 2n+1 }{ 6 } -\frac { n+1 }{ 4 } ) } \)
\(=\sqrt { (n+1)(\frac { 4n+2-3n-3 }{ 12 } ) } \)
\(=\sqrt { \frac { (n+1)(n-1) }{ 12 } } =\sqrt { \frac { { n }^{ 2 }-1 }{ 12 } } \)
\(\therefore \ Variance={ (SD) }^{ 2 }=\frac { { n }^{ 2 }-1 }{ 12 } \) -
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 \(\bar { x } +MD\left( \bar { x } \right) \) -
Let us make the following table from the given data.
Class fi cf Mid value (xi) |xi - M| fi|xi - M| 15.5 - 20.5 5 5 18 |18 - 38| = 20 100 20.5 - 25.5 6 11 23 |23 - 38| = 15 90 25.5 - 30.5 12 23 28 |28 - 38| = 10 120 30.5 - 35.5 14 37 33 |33 - 38| = 5 70 35.5 - 40.5 26 63 38 |38 - 38| = 0 0 40.5 - 45.5 12 75 43 |43 - 38| = 5 60 45.5 - 50.5 16 91 48 |48 - 38| = 10 160 50.5 - 55.5 9 100 53 |53 - 38| = 15 135 Total N = 100 735 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.
