Class 11 Maths NCERT Solutions for Chapter 15 Statistics Miscellaneous Exercise
Statistics Miscellaneous Exercise Solutions
Let the remaining two observations be x and y.
Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.
⇒ 60 + x+ y = 72
⇒ x + y = 12 ...(1)
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.
2. 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.
Solution
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 y_{i} = 3x, i.e., x_{i} = (1/3) y_{i}, for i = 1 to 6.
Therefore, variance of new observations = (1/6) × 864 = 144
Hence, the standard deviation of new observations is √144 = 12
(i) If wrong item is omitted.
(ii) If it is replaced by 12.
Incorrect mean = 10
Incorrect standard deviation = 2
That is, incorrect sum of observations = 200
Correct sum of observations = 200 - 8 = 192
Correct mean = (Correct sum )/19 = 192/19 = 10.1
= √4.09
= 2.02
Incorrect sum of observations = 200
Correct sum of observations = 200 - 8 + 12 = 204
∴ Correct mean = (Correct sum)/20 = 204/20 = 10.2
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?
Solution
Standard deviation of Mathematics = 12
Standard deviation of Physics = 15
Standard deviation of Chemistry = 20
The coefficient of variation (C.V.) is given by (Standard deviation )/Mean × 100
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.
7. 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.
Solution
Number of observations (n) = 100
Incorrect mean x = 20
Incorrect standard deviation (Ïƒ) = 3
Incorrect sum of observations = 2000
Correct sum of observations = 2000 - 21 - 21 - 18 = 2000 - 60 = 1940
∴ Correct mean = (correct sum)/(100 - 3) = 1940/97 = 20