Statistics
Category : 10th Class
Statistics
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 \[\overline{x}\,\,=a\,\,+\,\,\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h\] |
Mode= \[I+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\,\times \,\,h\]
Median = \[I+\left( \frac{\frac{n}{2}-cf}{f} \right)\,\times \,\,h\]
\[Mode=3\text{ }\left( Median \right)\text{ }-\text{ }2\text{ }\left( Mean \right)\]
Snap Test
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, \[{{f}_{1}}=\text{ }28\text{ }and\text{ }{{f}_{2}}=\text{ }\left( 52\text{ }-\text{ }28 \right)\text{ }=\text{ }24.\]
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, \[p\text{ }=\text{ }16.\]
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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