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Published on: 21/10/2019
Measures of Central Tendency - Arithmetic Mean
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1.
A students scored 36, 40, 42, 45 and 25 marks in Hindi, English, Mathematics, Social science and Science respectively. It was decided to give double weightage to the marks of Hindi and English and triple weightage to the marks of Science. Find weighted as well as simple mean.
2.
Average marks of 50 students were 42. Later it was decided to give a grace of 2 marks to al students. Marks of all students were adjusted accordingly. Find new mean.
3.
In a class average marks of boys and girls are 65 and 75 respectively. The combined average of all students taken together is 72. Find percentage of boys and girls in the class.
4.
There are 80 students in a class whose average marks in Statistics are 65. There are two sections in the class. In one section there are fifty students whose average marks are 62. Find the average marks of other section.
5.
Find out combined mean from the following data.
| Group A | Group B | Group C | |
|---|---|---|---|
| Mean | 12 | 20 | 10 |
| Number of items | 20 | 30 | 40 |
6.
From the following data of the marks obtained by 60 students of a class mean was found to be 41 Show that \(\sum _{ i=1 }^{ n }{ F({ x }_{ i }-\overline { x } ) } =0\)
| Marks | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|
| No of students | 8 | 12 | 20 | 10 | 6 | 4 |
7.
When is Arithmetic mean and weighted mean equal?
8.
Mention merits and demerits of Mean. And list out any three examples of 'scope of averages in your classroom.
9.
"Averages are indicators of a series". How?
10.
What is weighted mean? When is it useful?
11.
What are the objectives of a statistical average?
12.
Find Mean from the data given below:
| Mid Value | 3 | 9 | 15 | 21 | 27 | 33 | 30 |
| Frequency | 14 | 12 | 21 | 22 | 15 | 10 | 6 |
13.
If the average of the following data is 20.2, find the value of M:
| x | 10 | 15 | 20 | 25 | 30 |
| f | 6 | 8 | M | 10 | 6 |
14.
Calculate Mean from the data given below using step deviation method:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No of students | 8 | 12 | 20 | 10 | 6 | 4 |
15.
Find the average of the following data:
| Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
| No of students | 8 | 12 | 20 | 10 | 6 | 4 |
1.
| Subject | Weights | Marks | WX |
|---|---|---|---|
| Hindi | 2 | 36 | 72 |
| English | 2 | 40 | 80 |
| Mathematics | 1 | 42 | 42 |
| Social Science | 1 | 45 | 45 |
| Science | 3 | 25 | 75 |
| Total | \(\sum\)W=9 | \(\sum\)X = 198 | \(\sum\)WX = 314 |
Weighted Mean = \(\sum\)WX/\(\sum\)W = 314/9 = 34.88
Simple Mean = \(\sum\)X/N = 198/5 = 39.6
2.
Old \(\sum\) FX = 50 x 42 = 2100
New \(\sum\)FX = 2100 + (50 x 2) = 2200
New Mean = 2200/50 = 44
3.
| Boys | Girls | Combined | |
|---|---|---|---|
| Mean | 65 | 75 | 72 |
| Number of items | X | 100 - X | 100 |
\(\overline { x } 12=\frac { { N }_{ 1 }\overline { x } 1+{ N }_{ 2 }\overline { x } 2 }{ { N }_{ 1 }+{ N }_{ 2 } } \)
72 = \(\frac { \{ (65\times X)+(75\times 100-X)\} }{ 100 } \)
7200 = 65X + 7500 - 75X
10X = 300
100 - X = 70
There are 30% and 70% girls in the class.
4.
| Section A | Section B | Combined | |
|---|---|---|---|
| Mean | 62 | ? | 65 |
| Number of items | 50 | 30 | 80 |
\(\overline { x } 12=\frac { { N }_{ 1 }\overline { x } 1+{ N }_{ 2 }\overline { x } 2 }{ { N }_{ 1 }+{ N }_{ 2 } } \)
65 = \(\frac { (62+50)+(Y\times 30) }{ 80 } \)
30Y = 2494
Y = 2490/30 = 83
5.
Combined Mean
\(\overline { x } 123=\frac { { N }_{ 1 }\overline { x1 } +{ N }_{ 2 }\overline { x2 } +{ N }_{ 2 }\overline { x3 } }{ { N }_{ 1 }+{ N }_{ 2 }+{ N }_{ 3 } } \)
Combined Mean = \(\frac { (12\times 20)+(20\times 30)+(10\times 40) }{ 20+30+40 } \)
6.
| Marks (x) | No. of students (f) | \(({ x }_{ i }-\overline { x } )=0\) | \(F({ x }_{ i }-\overline { x } )=0\) |
|---|---|---|---|
| 20 | 8 | -21 | -168 |
| 30 | 12 | -11 | -132 |
| 40 | 20 | -1 | -20 |
| 50 | 10 | +9 | +90 |
| 60 | 6 | +19 | +114 |
| 70 | 4 | +29 | +116 |
| N = 60 | \(\sum _{ i=1 }^{ n }{ F({ x }_{ i }-\overline { x } ) } =0\) |
\(\overline { x } =\frac { 2460 }{ 60 } =41\)
7.
If the weights of all the observations are equal i.e. wI = w2 = w3..... = wn = W then the weighted A.M is equal to simple A.M
i.e. \(\overline {x}\)w = \(\overline {x}\)
8.
Merits of Arithmetic Mean:
(a) Arithmetic mean is most popular among averages used in statistical analysis.
(b) It is very simple to understand and easy to calculate.
(c) The calculation of A.M is based on all the observations in the series.
(d) The A.M is responsible for further algebraic treatment.
(e) It is strictly defined.
(j) It provides a good means of comparison.
(g) It has more sampling stability
Demerits of Arithmetic Mean:
(a) The A.M is affected by the extreme values in a series.
(b) In case of a missing observation in a series it is not possible to calculate the A.M.
(c) In case frequency distribution with open end classes the calculation of A.M is theoretically impossible.
(d) The arithmetic mean is an unsuitable average for qualitative data
9.
Averages show one figure which is representative of the entire series. It is simply impossible to remember income of each and every Indian. But it si quite easy to remember per capita income of India. A teacher cannot remember marks of all students in the class but she can remember mean or median marks in the class. Hence, it is rightly said that averages are indicators fo a series
10.
Weighted Mean is an average computed by giving different weights to some of the individual values. Basically you use a weighted mean when the data holds different "weights"/group sizes i.e. if you were looking for the mean of the number of people to visit a shop during its opening hours and you were told x went in the morning and y in the afternoon but the shop was open 3 hours in the morning and 6 in the evening just adding the two and dividing by 2 would not give you the correct average per hour. So you would need to weight the means i.e. (3x + 6y)/9 would be you actual average per hour as opposed to (x + y)/2.
11.
Averages occupy a prime place in the theory of statistical methods. That is why Bowley remarked, "Statistics is a science of averages." The following are them I objectives of an average:
1. Facilitates Comparison: The foremost purpose of average is that it facilitates comparison. For instance, a comparison of the production of jute in Maharashtra a Punjab shows that production of jute in Maharashtra is much more as compared Punjab.
2. Formulation of Policies: Averages are of great use in the formulation of various policy measures. For instance, when the Government finds that there is a fear of low product of sugar, it can formulate various policies to compensate the same.
3. Short Description: Averages help to present the raw data in a brief a systematic manner.
4. Representation of Universe: Average represents universe. According conclusions can be drawn in respect of the universe as a whole.
5. To represent huge mass of data in a summarized manner: It is difficult for a human being to grasp a large mass of data in mind bt an average summarizes such mass data into a single figure which is easier to understand and remember. None can remember income of all Indians but it is an easy task to remember per capita income of India.
12.
| Mid value(M) | Frequency(f) | FM |
| 3 | 14 | 42 |
| 9 | 12 | 108 |
| 15 | 21 | 315 |
| 21 | 22 | 462 |
| 27 | 16 | 405 |
| 33 | 10 | 330 |
| 30 | 6 | 180 |
| Total | \(\sum F\) =100 | \(\sum FX\)=1842 |
13.
| X | f | fx |
| 10 | 6 | 60 |
| 15 | 8 | 120 |
| 20 | M | 20M |
| 25 | 10 | 250 |
| 30 | 6 | 180 |
| 30+M | 610+20M |
\(\overline { x } =\frac { \sum { fX } }{ f } \)
20.2 =( 610 + 20M) / ( 30 + M)
20.2 ( 30 + M ) =610 + 20M
606 + 20.2M =610 + 20M
20.2M - 20M =610 - 606
0.2M= 4
M =4/0.2 =20
14.
Calculation of Arithmetic mean by direct method
| Marks | Frequency(f) | Mid value(M) | d=M-A(A=35) | d'=d/10 | Frequency* Midvalue(FM) |
| 0-10 | 8 | 5 | -30 | -3 | -240 |
| 10-20 | 12 | 15 | -20 | -2 | -240 |
| 20-30 | 20 | 25 | -10 | -1 | -200 |
| 30-40 | 10 | 35 | 0 | 0 | 0 |
| 40-50 | 6 | 45 | +10 | 1 | +60 |
| 50-60 | 4 | 55 | +20 | +2 | +80 |
| Total | \(\sum f\) =60 | \(\sum Fd\)=-54 |
\(\overline { x } =A+\frac { \sum _{ i=1 }^{ n }{ { f }_{ i } } { d }_{ i } }{ \sum _{ i=1 }^{ n }{ { f }_{ i } } } \times c\)
\(\overline{x}=35+\frac{-54}{60}\times10\) =25
15.
First find the class mark, Class Mark = (upper-class limit + Lower class limit) /2
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