10th Class Mathematics Statistics

 

Statistics

 

 

1. Mean of grouped data: It is the sum of the values of all the observations divided by the total number of observations.

 

2. The mean for grouped data can be found by

           

i. The direct method \[\overline{x}\,\,=\,\,\frac{\sum{{{f}_{i}}{{x}_{i}}}}{\sum{{{f}_{i}}}}\]

ii. The assumed mean method =\[\overline{x}\,\,=a\,\,+\,\,\frac{\sum{{{f}_{i}}{{d}_{i}}}}{\sum{{{f}_{i}}}}\]

iii. The step deviation method \[\overline{x}\,\,=a\,\,+\,\,\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h\]

 

3. Mode: It is the variate which occurs most often, i.e. which has the maximum frequency.

 

4. The mode for grouped data can be found using the formula

 

            Mode= \[I+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\,\times \,\,h\]

 

5. Median: The median is a measure of the central tendency which gives the value of the middlemost observation in the data.

 

6. The median for grouped data can be found using the formula:

 

            Median = \[I+\left( \frac{\frac{n}{2}-cf}{f} \right)\,\times \,\,h\]

 

7. Relationship among Mean, Median and Mode

            \[Mode=3\text{ }\left( Median \right)\text{ }-\text{ }2\text{ }\left( Mean \right)\]

 

 

Snap Test

 

 

 

1. The mean of the following frequency distribution is 62.8. Find the missing frequency x.

           

Class	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	5	8	x	12	7	8

 

           

(a) 15                            (b) 20

            (c) 10                            (d) 30

            (e) None of these

Ans.     (c)

            Explanation: According to the question:

                       

Class interval	Frequency fi	Mid-value xi	(fi \[\mathbf{\times }\] xi)
0 - 20	5	10	50
20 - 40	8	30	240
40 - 60	X	50	50x
60 - 80	12	70	840
80 - 100	7	90	630
100 - 120	8	110	880
	\[\sum{{{f}_{i}}\,\,=\,(40\,\,+\,\,x)}\]		\[({{f}_{i}}\,\,\times \,\,{{x}_{i}})\,=\,(2640\,\,+\,\,50x)\]

 

\[\therefore \]      \[Mean\text{ }\overline{x}\,\,=\,\,\frac{\sum{({{f}_{i}}\,\times \,{{x}_{i}})}}{\sum{{{f}_{i}}}}\]                     \[\Rightarrow \,\,\,\,\,\,\frac{2640\,+50x}{40\,+\,x}\,\,=\,\,62.8\]

\[\Rightarrow \,\,\,\,\,\,\frac{2640\,+50x}{40\,+\,x}\,\,=\,\,\frac{628}{10}\]

\[\Rightarrow \,\,\,\,\,\,\,\,26400\,\,+\,\,500x\,\,=\,\,25120\,\,+\,\,628x\]

\[\Rightarrow \,\,\,\,\,\,\,\,128x\,\,=\,\,1280\,\,\,\,\,\,\Rightarrow \,\,\,x\,\,=\,\,10\]

 

2.         The mean of the following frequency table is 50. But the frequencies \[{{\mathbf{f}}_{\mathbf{1}}}\text{ }\mathbf{and}\text{ }{{\mathbf{f}}_{\mathbf{2}}}\] in class 20 – 40 and 60 – 80 are missing frequencies.

 

 

Class Interval	10 - 20	20 - 40	40 - 60	60 - 80	80 - 100	Total
	17	\[{{f}_{1}}\]	32	\[{{f}_{2}}\]	19	120

           

          

(a) 28, 24                      (b) 26, 24

            (c) 27, 20                       (d) 30, 28

            (e) None of these

Ans.     (a)

Explanation: According to the question: \[17\text{ }+\text{ }{{f}_{1}}+\text{ }32\text{ }+\text{ }{{f}_{2}}+\text{ }19\text{ }=\text{ }120\,\,\,\,\,\text{ }{{f}_{2}}=\text{ }\left( 52\text{ }\text{ }{{f}_{1}} \right).\]

            Now we may prepare the table given below:

           

Class interval	Frequency \[{{\mathbf{f}}_{\mathbf{i}}}\]	Class Mark \[{{\mathbf{x}}_{\mathbf{i}}}\]	\[\begin{array}{*{35}{l}}    \left( {{\mathbf{f}}_{\mathbf{i}}}\mathbf{\times }{{\mathbf{x}}_{\mathbf{i}}} \right)  \\ \end{array}\]
0 - 20	17	10	170
20 ? 40	\[{{f}_{1}}\]	30	\[30{{f}_{1}}\]
40 - 60	32	50	1600
60 - 80	\[52\text{ }-\text{ }{{f}_{1}}\]	70	\[3640\text{ }-\text{ }70{{f}_{1}}\]
80 - 100	19	90	1710
	\[\sum{{{f}_{i}}\,\,=\,\,120}\]		\[\sum{({{f}_{i}}\,\,\times \,\,{{x}_{i}})\,\,\,=\,\,\,(7120\,\,-\,\,40{{f}_{1}})}\]

 

\[\therefore \,\,\,\,\,\,\,\,Mean\,\,\overline{x}\,\,=\,\,\frac{\sum{({{f}_{i}}\times {{x}_{i}})}}{\sum{{{f}_{i}}}}\,\,=\,\,\frac{(7120-40{{f}_{1}})}{120}\,\,\,=\,\,\,\frac{(178-{{f}_{1}})}{3}\]

 

So, \[\frac{178-{{f}_{1}}}{3}\,\,=\,\,50\,\,\,\Rightarrow \,\,178-{{f}_{1}}\,\,=\,\,150\,\,\,\Rightarrow \,\,{{f}_{1}}=28\]

Thus, \[{{f}_{1}}=\text{ }28\text{ }and\text{ }{{f}_{2}}=\text{ }\left( 52\text{ }-\text{ }28 \right)\text{ }=\text{ }24.\]

 

3. Find the mean of the following data:

           

Class Interval	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	7	8	12	13	10

 

          

(a) 25.3                         (b) 27.2

            (c) 20.23                        (d) 26.1

            (e) None of these

Ans.     (b)

            Explanation: We have:

 

Class interval	Frequency \[{{\mathbf{f}}_{\mathbf{i}}}\]	Class Mark \[{{\mathbf{x}}_{\mathbf{i}}}\]	\[\begin{array}{*{35}{l}}    \left( {{\mathbf{f}}_{\mathbf{i}}}\,\,\mathbf{\times }\,\,{{\mathbf{x}}_{\mathbf{i}}} \right)  \\ \end{array}\]
0 - 10	7	5	35
10 - 20	8	15	120
20 - 30	12	25	300
30 - 40	13	35	455
40 - 50	10	45	450
	\[\sum{{{f}_{i}}\,\,=\,\,50}\]		\[\sum{({{f}_{i}}\,\,\times \,\,{{x}_{i}})\,\,=\,\,1360}\]

 

\[\therefore \,\,\,\,\,Mean\,\,=\,\,\frac{\sum{({{f}_{i}}\,\,\times \,\,{{x}_{i}})}}{\sum{{{f}_{i}}}}\,\,=\,\,\frac{1360}{50}\,\,=\,\,27.2\]

 

4.         The arithmetic mean of the following frequency distribution is 25. Determine the value of p

           

Class	0 - 10	10 - 20	20 - 30	30 - 40	40 - 50
Frequency	5	18	15	p	6

 

           

(a) 12                            (b) 14

            (c) 10                            (d) 16

            (e) None of these

Ans.     (d)

            Explanation: We have:

           

Class interval	Frequency	Mid-value	\[\left( {{\mathbf{f}}_{\mathbf{i}}}\,\,\mathbf{\times }\,\,{{\mathbf{x}}_{\mathbf{i}}} \right)\]
0 - 10	5	5	25
10 - 20	18	15	270
20 - 30	15	25	375
30 - 40	P	35	35p
40 - 50	6	45	270
	\[\sum{{{f}_{i}}\,\,=\,\,(44+p)}\]		\[\sum{({{f}_{i}}\,\,\times \,\,{{x}_{i}})\,\,=\,\,(940\,\,+\,\,35p)}\]

 

\[\therefore \]      Mean \[\overline{x}=\frac{\sum{({{f}_{i}}\,\,\times \,\,{{x}_{i}})}}{\sum{{{f}_{i}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\frac{(940+35p)}{(44+p)}\,\,\,=\,\,25\]

 

\[\Rightarrow \] \[\left( 940\text{ }+\text{ }35p \right)\text{ }=\text{ }25\left( 44\text{ }+\text{ }p \right)\]

                                                            \[\Rightarrow \] \[(35p-25p)\,\,=\,\,(1100-940)\]                                                   

\[\Rightarrow \] \[10p\text{ }=\text{ }160~~\Rightarrow \,\,\,~p\text{ }=\text{ }16.\]

            Hence, \[p\text{ }=\text{ }16.\]

 

5. The following table gives the marks scored by 50 students in class test:

           

Marks	0 - 100	100 - 200	200 - 300	300 - 400	400 - 500	500 - 600
No. of  students	2	8	12	20	5	3

           

           

Find the mean marks scored by a student in the class test.

            (a) 304                          (b) 306

            (c) 310                          (d) 308

            (e) None of these

Ans.     (a)

            Explanation: Here, \[h=\text{ }100\]. Thus, we have:

                       

Class interval	Frequency \[{{\mathbf{f}}_{\mathbf{i}}}\]	Mid-value \[{{\mathbf{x}}_{\mathbf{i}}}\]	\[{{\mathbf{u}}_{\mathbf{i}}}\mathbf{=}\frac{{{\mathbf{x}}_{\mathbf{i}}}\mathbf{-A}}{\mathbf{h}}\]   \[\mathbf{=}\,\,\frac{\mathbf{(}{{\mathbf{x}}_{\mathbf{i}}}\mathbf{-250)}}{\mathbf{100}}\]	\[\left( {{\mathbf{f}}_{\mathbf{i}}}\mathbf{\times }{{\mathbf{u}}_{\mathbf{i}}} \right)\]
0 - 100	2	50	-2	-4
100 - 200	8	150	-1	-8
200 - 300	12	\[250\text{ }=\text{ }A\]	0	0
300 - 400	20	350	1	20
400 - 500	5	450	2	10
500 - 600	3	550	3	9
	\[\sum{{{f}_{i}}\,\,=\,\,50}\]			\[\sum{({{f}_{i}}\,\times \,{{u}_{i}})}\,=\,27\]

 

 

Thus, A             = \[250,\,\,\,h\,\,=\,\,100,\,\sum{{{f}_{1}}=50,\,\sum{({{f}_{i}}\,\,\times \,\,{{u}_{i}})=27.}}\]

\[\therefore \,\,\,\,Mean,\,\,\overline{x}\,\,\,\,\,\,\,\,\,\,\,\,=\,\,A\,+\,\left( h\times \,\,\frac{\sum{({{f}_{i}}\,\,\times \,\,{{u}_{i}})}}{\sum{{{f}_{i}}}} \right)\]

\[=250+\left( 100\,\,\times \,\,\frac{27}{50} \right)\,\,\,=\,250\,\,+\,\,54\,\,=\,\,304\]

            \[Hence,\text{ }mean\text{ }marks=\text{ }304.\]