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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.
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
x2 + y2 + 2xy = 144 …(3)
From (2) and (3), we obtain
2xy = 64 … (4)
Subtracting (4) from (2), we obtain
x2 + y2 – 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.
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.
Let the remaining two observations be x and y.
The observations are 2, 4, 10, 12, 14, x, y.
From (1), we obtain
x2 + y2 + 2xy = 196 … (3)
From (2) and (3), we obtain
2xy = 196 – 100
⇒ 2xy = 96 … (4)
Subtracting (4) from (2), we obtain
x2 + y2 – 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.
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.
Let the observations be x1, x2, x3, x4, x5, and x6.
It is given that mean is 8 and standard deviation is 4.
If each observation is multiplied by 3 and the resulting observations are yi, then
From (1) and (2), it can be observed that,
Substituting the values of xi and 
in (2), we obtain
Therefore, variance of new observations = 
Hence, the standard deviation of new observations is 
Given that 
is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the mean and variance of the observations ax1, ax2, ax3 …axn are 
and a2 σ2, respectively (a ≠ 0).
The given n observations are x1, x2 … xn.
Mean =
Variance = σ2
If each observation is multiplied by a and the new observations are yi, then
Therefore, mean of the observations, ax1, ax2 … axn, is 
.
Substituting the values of xiand in (1), we obtain
Thus, the variance of the observations, ax1, ax2 … axn, is a2 σ2.
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.
(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
The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:
|
Subject |
Mathematics |
Physics |
Chemistry |
|
Mean |
42 |
32 |
40.9 |
|
Standard deviation |
12 |
15 |
20 |
Which of the three subjects shows the highest variability in marks and which shows the lowest?
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.
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.
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
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