Class 10 Mathematics Note

Statistics

Statistics

Mean

A mean is defined as the average of the numbers. The data may be individual, discrete and continuous .The method of calculating the means depend upon the nature of data. It is denoted byx̅

For individual data:x̅ =$\frac{{{{\rm{x}}_1} + {{\rm{x}}_2} + {{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots  \ldots {{\rm{x}}_{\rm{n}}}}}{{\rm{N}}}$=$\frac{{\Sigma {\rm{x}}}}{{\rm{N}}}{\rm{\: }}$

For discrete data: x̅ =$\frac{{{{\rm{f}}_1}{{\rm{x}}_1} + {{\rm{f}}_2}{{\rm{x}}_2} + {{\rm{f}}_3}{{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots  \ldots {{\rm{f}}_{\rm{n}}}{{\rm{x}}_{\rm{n}}}}}{{{{\rm{f}}_1} + {\rm{\: }}{{\rm{f}}_2} + {\rm{\: }}{{\rm{f}}_3} \ldots  \ldots  \ldots  \ldots ..{{\rm{f}}_{\rm{n}}}}}$$ = {\rm{\: }}\frac{{\Sigma {\rm{fx}}}}{{\rm{N}}}$

For continuous data: $\frac{{{{\rm{f}}_1}{{\rm{x}}_1} + {{\rm{f}}_2}{{\rm{x}}_2} + {{\rm{f}}_3}{{\rm{x}}_3} + {\rm{\: }}{{\rm{x}}_4} \ldots  \ldots {{\rm{f}}_{\rm{n}}}{{\rm{x}}_{\rm{n}}}}}{{{{\rm{f}}_1} + {\rm{\: }}{{\rm{f}}_2} + {\rm{\: }}{{\rm{f}}_3} \ldots  \ldots  \ldots  \ldots ..{{\rm{f}}_{\rm{n}}}}}$=$\frac{{\Sigma {\rm{fx}}}}{{\rm{N}}}$

 

 

Example 1

In a grouped data Σ fm = 25a + 150, mean x̅ = 20 and N = 10+ a .Find of a .

Soln

x̅ = 20

N = 10+ a

Σ fm = 25a +150

We have

x̅ = $\Sigma \frac{{{\rm{fm}}}}{{\rm{N}}}$= $\frac{{25{\rm{a\: }} + 150}}{{10 + {\rm{\: a\: }}}}$

$\frac{{25{\rm{a\: }} + 150}}{{10 + {\rm{\: a\: }}}}$=20

25a +150 = 200 +20a

5a =50

a= 10

Example 2

If the continuous series, the mean x̅ =24, Σ fm = 400 and the number of terms N = 6 + $\frac{{\rm{m}}}{9}$ .Find the value of m

soln

x̅ = 45

N = 6 + $\frac{{\rm{m}}}{9}$

Σ fm = 400

We have

x̅ = $\Sigma \frac{{{\rm{fm}}}}{{\rm{N}}}$= $\frac{{400}}{{{\rm{\: }}6{\rm{\: }} + {\rm{\: }}\frac{{\rm{m}}}{9}}}$

$\frac{{400}}{{{\rm{\: }}6{\rm{\: }} + {\rm{\: }}\frac{{\rm{m}}}{9}}}$= 45

5m = 400-270

m =26

 

Median and quartile

Median is the mid –value of the data.

 

For the individual and discrete data

Median = value of $\frac{{{\rm{N}} + 1}}{2}$ item if the number of the terms is odd

= average value of$\frac{{{\rm{\: N}}}}{2}$${\rm{and\: }}\frac{{{\rm{N}} + 2}}{2}$item if the number of the terms is even

For continuous data

Median =L + $\frac{{\frac{{\rm{N}}}{2} - {\rm{c}}.{\rm{f}}}}{{\rm{f}}}$ *i

N= total number of items

L = lower limit of median class

C.F = C.F of the preceding the median class

F= frequency of median class

I = class interval

 

Examples 3

The following marks are obtained by students in mathematics in an examination

51,20,60,34,48,61,79,25,39,52,43,46,37,40,67,76,71,33,44,55

a. Make frequency table of class- interval 10

b. find the median

 

soln

Marksobtained

frequency

CF

20 – 30

1

1

30 – 40

4

5

40 – 50

5

11

50-60

3

14

60- 70

3

17

70-80

3

20

 

N = 20

 

 

Now,

Md = ${\frac{{20}}{2}^{{\rm{th}}}}$item

= ${10^{{\rm{th}}}}$item

= 40-50 th item

So, the cf = 6 . So, the class is 40-50

Now, L = 40, h = 10, f = 5 cf = 6

So, Md = $40 + \frac{{10 - 5}}{5}{\rm{*}}10$

= 50

 

Example 2

If the median of the following datais 24 .find the value y.

soln

Marks obtained

Frequency

CF

0 -10

4

4

10 – 20

12

16

20 – 30

y

16+ y

30-40

9

25+ y

40 – 50

5

30+ y

 

N=30 + y

 

 

Now,

Md = ${\frac{{\rm{N}}}{2}^{{\rm{th}}}}$item

$\frac{{\rm{N}}}{2}$= ${\frac{{{\rm{\: }}30 + {\rm{y\: }}}}{2}^{{\rm{th}}}}$item

 Median = 24 median lies between 20 -30

Now, L = 20, h = 10, f = y, cf = 16

So, Md = $20 + \frac{{\frac{{\rm{N}}}{2} - {\rm{cf}}}}{{\rm{f}}}{\rm{*}}10$

24 = $20 + \frac{{{\rm{\: }}\frac{{{\rm{\: }}30 + {\rm{y\: }}}}{2}{\rm{\: }} - {\rm{y\: }}}}{{\rm{y}}}{\rm{*}}10$

Y= 10

 

Quartiles

A quartile is a type of quantile. The first quartile (Q1) is defined as the middle number between the smallest number and the median of the data set. The second quartile (Q2) is the median of the data. The third quartile (Q3) is the middle value between the median and the highest value of the data set.

Q1 =L + $\frac{{\frac{{\rm{N}}}{4} - {\rm{c}}.{\rm{f}}}}{{\rm{f}}}$ *i

Q2= L + $\frac{{\frac{{\rm{N}}}{2} - {\rm{c}}.{\rm{f}}}}{{\rm{f}}}$ *i

Q3= L + $\frac{{\frac{{3{\rm{N}}}}{4} - {\rm{c}}.{\rm{f}}}}{{\rm{f}}}$ *i

N= total number of items

L = lower limit of median class

C.F = C.F of the preceding the median class

F= frequency of median class

I = class interval

 

Example 4

Find the Q3 from the following data

Marks obtained

Frequency

10-20

3

20-30

5

30-40

4

40-50

5

50-60

4

60-70

2

70-80

3

 

 

Example 5

Soln

Marks obtained

Frequency

CF

10-20

3

3

20-30

5

8

30-40

4

12

40-50

5

17

50-60

4

21

60-70

2

23

70-80

3

26

 

N= 26

 

 

Now,

Q3 = ${\frac{{3{\rm{N}}}}{4}^{{\rm{th}}}}$item

= ${\frac{{3{\rm{*}}26}}{4}^{{\rm{th}}}}$item

=19.5 th item

So, the class is 50-60

Now, L = 50, h = 10, f = 4, cf = 17

So, Q3 = $50 + \frac{{\frac{{3{\rm{N}}}}{4} - {\rm{cf}}}}{{\rm{f}}}{\rm{*h}}$

= $50 + \frac{{19.5 - 17}}{4}{\rm{*}}10$

 = 56.25

 

Example 5

 Find Q3 from the following data

Soln

Marks obtained

Frequency

CF

10-20

3

3

20-30

5

8

30-40

4

12

40-50

5

17

50-60

4

21

60-70

2

23

70-80

3

26

 

N= 26

 

 

Now,

Q3 = ${\frac{{3{\rm{N}}}}{4}^{{\rm{th}}}}$item

= ${\frac{{3{\rm{*}}26}}{4}^{{\rm{th}}}}$item

=19.5 th item

So, the class is 50-60

Now, L = 50, h = 10, f = 4, cf = 17

So, Q3 = $50 + \frac{{\frac{{3{\rm{N}}}}{4} - {\rm{cf}}}}{{\rm{f}}}{\rm{*h}}$

= $50 + \frac{{19.5 - 17}}{4}{\rm{*}}10$

 = 56.25

 


Go Top