Q2: Find the mean deviation about the mean for the data 38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Ans: The given data is
38, 70, 48, 40, 42, 55, 63, 46, 54, 44
Mean of the given data,
The deviations of the respective observations from the mean are
–12, 20, –2, –10, –8, 5, 13, –4, 4, –6
The absolute values of the deviations, i.e. , are
12, 20, 2, 10, 8, 5, 13, 4, 4, 6
The required mean deviation about the mean is
Q3: Find the mean deviation about the median for the data.
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Ans: The given data is
13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
Here, the numbers of observations are 12, which is even.
Arranging the data in ascending order, we obtain
10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18
The deviations of the respective observations from the median, i.e. are
–3.5, –2.5, –2.5, –1.5, –0.5, –0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The absolute values of the deviations, , are
3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5
The required mean deviation about the median is
Q4: Find the mean deviation about the median for the data
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Ans: The given data is
36, 72, 46, 42, 60, 45, 53, 46, 51, 49
Here, the number of observations is 10, which is even.
Arranging the data in ascending order, we obtain
36, 42, 45, 46, 46, 49, 51, 53, 60, 72
The deviations of the respective observations from the median, i.e. are
–11.5, –5.5, –2.5, –1.5, –1.5, 1.5, 3.5, 5.5, 12.5, 24.5
The absolute values of the deviations, , are
11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
Thus, the required mean deviation about the median is
Q5: Find the mean deviation about the mean for the data.
Ans:
Q6: Find the mean deviation about the mean for the data
Ans:
Q7: Find the mean deviation about the median for the data.
Ans: The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
Here, N = 26, which is even.
Median is the mean of 13^{th} and 14^{th} observations. Both of these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.
The absolute values of the deviations from median, i.e. are
and
Q8: Find the mean deviation about the median for the data
Ans: The given observations are already in ascending order.
Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.
Here, N = 29, which is odd.
observation = 15^{th} observation
This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
∴ Median = 30
The absolute values of the deviations from median, i.e. are
∴
Q9: Find the mean deviation about the mean for the data.
Ans: The following table is formed
Here,
Q10: Find the mean deviation about the mean for the data
Ans: The following table is formed.
Here,
Q11: Find the mean deviation about median for the following data:
Ans: The following table is formed.
The class interval containing the or 25^{th} item is 20 – 30.
Therefore, 20 – 30 is the median class.
It is known that,
Here, l = 20, C = 14, f = 14, h = 10, and N = 50
∴ Median =
Thus, mean deviation about the median is given by,
Q12: Calculate the mean deviation about median age for the age distribution of 100 persons given below:
Ans: The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
The table is formed as follows.
Age  Number f_{i}  Cumulative frequency (c.f.)  Midpoint x_{i}  x_{i} – Med.  f_{i} x_{i} – Med. 
15.520.5  5  5  18  20  100 
20.525.5  6  11  23  15  90 
25.530.5  12  23  28  10  120 
30.535.5  14  37  33  5  70 
35.540.5  26  63  38  0  0 
40.545.5  12  75  43  5  60 
45.550.5  16  91  48  10  160 
50.555.5  9  100  53  15  135 
 100 


 735 
The class interval containing the or 50^{th} item is 35.5 – 40.5.
Therefore, 35.5 – 40.5 is the median class.
It is known that,
Here, l = 35.5, C = 37, f = 26, h = 5, and N = 100
Thus, mean deviation about the median is given by,
Q1: Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
Ans: 6, 7, 10, 12, 13, 4, 8, 12
Mean,
The following table is obtained.
Q2: Find the mean and variance for the first n natural numbers
Ans: The mean of first n natural numbers is calculated as follows.
Q3: Find the mean and variance for the first 10 multiples of 3
Ans: The first 10 multiples of 3 are
3, 6, 9, 12, 15, 18, 21, 24, 27, 30
Here, number of observations, n = 10
The following table is obtained.
Q4: Find the mean and variance for the data
Ans: The data is obtained in tabular form as follows.
x_{i}  f i  f_{i}x_{i}  
6  2  12  –13  169  338 
10  4  40  –9  81  324 
14  7  98  –5  25  175 
18  12  216  –1  1  12 
24  8  192  5  25  200 
28  4  112  9  81  324 
30  3  90  11  121  363 
40  760  1736 
Here, N = 40,
Q5: Find the mean and variance for the data
Ans: The data is obtained in tabular form as follows.
x_{i}  f i  f_{i}x_{i}  
92  3  276  –8  64  192 
93  2  186  –7  49  98 
97  3  291  –3  9  27 
98  2  196  –2  4  8 
102  6  612  2  4  24 
104  3  312  4  16  48 
109  3  327  9  81  243 
22  2200  640 
Here, N = 22,
Q6: Find the mean and standard deviation using shortcut method.
Ans: The data is obtained in tabular form as follows.
x_{i}  f_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
60  2  –4  16  –8  32 
61  1  –3  9  –3  9 
62  12  –2  4  –24  48 
63  29  –1  1  –29  29 
64  25  0  0  0  0 
65  12  1  1  12  12 
66  10  2  4  20  40 
67  4  3  9  12  36 
68  5  4  16  20  80 
100  220  0  286 
Mean,
Q7: Find the mean and variance for the following frequency distribution.
Classes  030  3060  6090  90120  120150  150180  180210 
Frequencies  2  3  5  10  3  5  2 
Ans:
Class  Frequency f_{i}  Midpoint x_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
030  2  15  –3  9  –6  18 
3060  3  45  –2  4  –6  12 
6090  5  75  –1  1  –5  5 
90120  10  105  0  0  0  0 
120150  3  135  1  1  3  3 
150180  5  165  2  4  10  20 
180210  2  195  3  9  6  18 
30  2  76 
Mean,
Q8: Find the mean and variance for the following frequency distribution.
Classes  010  1020  2030  3040  4050 
Frequencies  5  8  15  16  6 
Ans:
Class  Frequency f_{i}  Midpoint x_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
010  5  5  –2  4  –10  20 
1020  8  15  –1  1  –8  8 
2030  15  25  0  0  0  0 
3040  16  35  1  1  16  16 
4050  6  45  2  4  12  24 
50  10  68 
Mean,
Q9: Find the mean, variance and standard deviation using shortcut method
Ans:
Class Interval  Frequency f_{i}  Midpoint x_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
7075  3  72.5  –4  16  –12  48 
7580  4  77.5  –3  9  –12  36 
8085  7  82.5  –2  4  –14  28 
8590  7  87.5  –1  1  –7  7 
9095  15  92.5  0  0  0  0 
95100  9  97.5  1  1  9  9 
100105  6  102.5  2  4  12  24 
105110  6  107.5  3  9  18  54 
110115  3  112.5  4  16  12  48 
60  6  254 
Mean,
Q10: The diameters of circles (in mm) drawn in a design are given below:
Ans:
Class Interval  Frequency f_{i}  Midpoint x_{i}  f_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
32.536.5  15  34.5  –2  4  –30  60 
36.540.5  17  38.5  –1  1  –17  17 
40.544.5  21  42.5  0  0  0  0 
44.548.5  22  46.5  1  1  22  22 
48.552.5  25  50.5  2  4  50  100 
 100 


 25  199 
Here, N = 100, h = 4
Let the assumed mean, A, be 42.5.
Mean,
Old NCERT Questions
Q1: From the data given below state which group is more variable, A or B?
Marks  1020  2030  3040  4050  5060  6070  7080 
Group A  9  17  32  33  40  10  9 
Group B  10  20  30  25  43  15  7 
Ans: Firstly, the standard deviation of group A is calculated as follows.
Marks  Group A f_{i}  Midpoint x_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
1020  9  15  –3  9  –27  81 
2030  17  25  –2  4  –34  68 
3040  32  35  –1  1  –32  32 
4050  33  45  0  0  0  0 
5060  40  55  1  1  40  40 
6070  10  65  2  4  20  40 
7080  9  75  3  9  27  81 
150  –6  342 
Here, h = 10, N = 150, A = 45
The standard deviation of group B is calculated as follows.
Marks  Group B f_{i}  Midpoint x_{i}  y_{i}^{2}  f_{i}y_{i}  f_{i}y_{i}^{2}  
1020  10  15  –3  9  –30  90 
2030  20  25  –2  4  –40  80 
3040  30  35  –1  1  –30  30 
4050  25  45  0  0  0  0 
5060  43  55  1  1  43  43 
6070  15  65  2  4  30  60 
7080  7  75  3  9  21  63 
150  –6  366 
Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
Thus, group B has more variability in the marks.
Q2: From the prices of shares X and Y below, find out which is more stable in value:
X  35  54  52  53  56  58  52  50  51  49 
Y  108  107  105  105  106  107  104  103  104  101 
Ans: The prices of the shares X are
35, 54, 52, 53, 56, 58, 52, 50, 51, 49
Here, the number of observations, N = 10
The following table is obtained corresponding to shares X.
x_{i}  
35  –16  256 
54  3  9 
52  1  1 
53  2  4 
56  5  25 
58  7  49 
52  1  1 
50  –1  1 
51  0  0 
49  –2  4 
350 
The prices of share Y are
108, 107, 105, 105, 106, 107, 104, 103, 104, 101
The following table is obtained corresponding to shares Y.
y_{i}  
108  3  9 
107  2  4 
105  0  0 
105  0  0 
106  1  1 
107  2  4 
104  –1  1 
103  –2  4 
104  –1  1 
101  –4  16 
40 
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.
Q3: An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:
Firm A  Firm B  
No. of wage earners  586  648 
Mean of monthly wages  Rs 5253  Rs 5253 
Variance of the distribution of wages  100  121 
Ans: (i) Monthly wages of firm A = Rs 5253
Number of wage earners in firm A = 586
∴Total amount paid = Rs 5253 × 586
Monthly wages of firm B = Rs 5253
Number of wage earners in firm B = 648
∴Total amount paid = Rs 5253 × 648
Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
(ii) Variance of the distribution of wages in firm A = 100
∴ Standard deviation of the distribution of wages in firm
A ((σ_{1}) =
Variance of the distribution of wages in firm = 121
∴ Standard deviation of the distribution of wages in firm
The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.
Thus, firm B has greater variability in the individual wages.
Q4: The following is the record of goals scored by team A in a football session:
For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?
Ans: The mean and the standard deviation of goals scored by team A are calculated as follows.
No. of goals scored  No. of matches  f_{i}x_{i}  x_{i}^{2}  f_{i}x_{i}^{2} 
0  1  0  0  0 
1  9  9  1  9 
2  7  14  4  28 
3  5  15  9  45 
4  3  12  16  48 
25  50  130 
Thus, the mean of both the teams is same.
The standard deviation of team B is 1.25 goals.
The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.
Thus, team A is more consistent than team B.
Q5: The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:
Which is more varying, the length or weight?
Ans:
Here, N = 50
∴ Mean,
Mean,
Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.
Q6: The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.
Ans: Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
From (1), we obtain
x^{2} + y^{2} + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2} – 2xy = 80 – 64 = 16
⇒ x – y = ± 4 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 4, when x – y = 4
x = 4 and y = 8, when x – y = –4
Thus, the remaining observations are 4 and 8.
Q7: The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
Ans: Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.
From (1), we obtain
x^{2} + y^{2} + 2xy = 196 … (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 … (4)
Subtracting (4) from (2), we obtain
x^{2} + y^{2} – 2xy = 100 – 96
⇒ (x – y)^{2} = 4
⇒ x – y = ± 2 … (5)
Therefore, from (1) and (5), we obtain
x = 8 and y = 6 when x – y = 2
x = 6 and y = 8 when x – y = – 2
Thus, the remaining observations are 6 and 8.
Q8: The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
Ans: Let the observations be x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, and x_{6}.
It is given that mean is 8 and standard deviation is 4.
If each observation is multiplied by 3 and the resulting observations are y_{i}, then
From (1) and (2), it can be observed that,
Substituting the values of x_{i} and in (2), we obtain
Therefore, variance of new observations =
Hence, the standard deviation of new observations is
Q9: Given that is the mean and σ^{2} is the variance of n observations x_{1}, x_{2} … x_{n}. Prove that the mean and variance of the observations ax_{1}, ax_{2}, ax_{3} …ax_{n} are and a^{2} σ^{2}, respectively (a ≠ 0).
Ans: The given n observations are x_{1}, x_{2} … x_{n}.
Mean =
Variance = σ^{2}
If each observation is multiplied by a and the new observations are y_{i}, then
Therefore, mean of the observations, ax_{1}, ax_{2} … ax_{n}, is .
Substituting the values of x_{i}and in (1), we obtain
Thus, the variance of the observations, ax_{1}, ax_{2} … ax_{n}, is a^{2} σ^{2}.
Q10: The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Ans: (i) Number of observations (n) = 20
Incorrect mean = 10
Incorrect standard deviation = 2
That is, incorrect sum of observations = 200
Correct sum of observations = 200 – 8 = 192
∴ Correct mean
(ii) When 8 is replaced by 12,
Incorrect sum of observations = 200
∴ Correct sum of observations = 200 – 8 + 12 = 204
Q11: The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
Which of the three subjects shows the highest variability in marks and which shows the lowest?
Ans: Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) is given by .
The subject with greater C.V. is more variable than others.
Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.
Q12: The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
Ans: Number of observations (n) = 100
Incorrect mean
Incorrect standard deviation
∴ Incorrect sum of observations = 2000
⇒ Correct sum of observations = 2000 – 21 – 21 – 18 = 2000 – 60 = 1940
Standered deviation
Correct Standered deviation
209 videos443 docs143 tests

1. What are the different types of statistics? 
2. How is central tendency calculated in statistics? 
3. What is the importance of statistics in research? 
4. How can outliers affect statistical analysis? 
5. What are the different measures of variability in statistics? 
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