# 10th Class Mathematics Statistics

Statistics

Category : 10th Class

Statistics

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

1. 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 $\overline{x}\,\,=a\,\,+\,\,\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h$

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

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

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

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

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

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

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

1. 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, $p\text{ }=\text{ }16.$

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

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

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

$Hence,\text{ }mean\text{ }marks=\text{ }304.$

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