Solution : According to given condition, 6.80 = \((6-a)^2 + (6-b)^2 + (6-8)^2 + (6-5)^2 + (6-10)^2\over 5\) \(\implies\) 34 = \((6-a)^2 + (6-b)^2\) + 4 + 1 + 16 \(\implies\) \((6-a)^2 + (6-b)^2\) = 13 \(\implies\) \((6-a)^2 + (6-b)^2\) = 13 = \(3^2\) + \(2^2\) \(\implies\) a = 3 and b = 4 Similar […]
Solution : \(\therefore\) Variance = [\({1\over n}\sum{(x_i)^2}\)] – \((\bar{x})^2\) = \(1\over n\)[\(2^2 + 4^2 + ….. + (2n)^2\)] – \((n+1)^2\) = \(1\over n\)\(2^2 [ 1^2 + 2^2 + ….. + n^2]\) – \((n+1)^2\) = \(4\over n\) \(n(n + 1)(2n + 1)\over 6\) – \((n+1)^2\) = \((n + 1)[(2n + 1) – 3(n + 1)\over 3\)
Find the variance of first n even natural numbers ? Read More »
Solution : Correct mean = old mean + 2 = 30 + 2 = 32 As standard deviation is independent of change of origin. Hence, it remains same. Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is The median
Solution : Median of a, 2a, 3a, 4a, ….. . 50a is \(25a + 26a\over 2\) = 25.5a Mean deviation = \(\sum{|x_i – Median|}\over N\) \(\implies\) 50 = \(1\over 50\) {2|a|.(0.5 + 1.5 + …… + 24.5)] \(\implies\) 2500 = 2|a|. \(25\over 2\) (25) \(\implies\) |a| = 4 Similar Questions The mean and variance of
Question : All the students of a class performed poorly in mathematics. The teacher decided to give grace marks of 10 to each of the students. Which statistical measure will not change even after the grace marks were given ? (a) Mean (b) Median (c) Mode (d) Variance Solution : If initially all marks were
Solution : \({\sigma^2}\) = \(\sum(x_i – \bar{x})^2\over n\) \(\bar{X}\) = \(\sum x_i\over n\) = \(2 + 4 + 6 + 8 + ….. + 100\over 50\) = 51 \({\sigma^2}\) = \(2^2 + 4^2 + …. + 100^2\over 50\) – \(51^2\) = 833 Similar Questions The mean and variance of a random variable X having a
The variance of first 50 even natural numbers is Read More »
Solution : As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then \({\sigma}^2\) = \(\sum{(x_i – \bar{x})}^2\over n\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9 …..(1) Also \(\bar{x}\) = 4 \(\implies\)
Solution : Total marks obtained from three subjects out of 300 = 75 + 80 + 85 = 240 if the marks of another subject is added then the total marks obtained out of 400 is greater than 240 if marks obtained in fourth subject is 0 then minimum average marks = \(240\over 400\)\(\times\)100 =
Solution : As given \(\bar{x}\) = \(x_1 + x_2 + …. + x_n\over n\) If the mean of the series \(x_i\) + 2i, i = 1, 2, ….., n be \(\bar{X}\), then \(\bar{X}\) = \((x_1+2) + (x_2+2.2) + (x_3+2.3) + …. + (x_n + 2.n)\over n\) = \(x_1 + x_2 + …. + x_n\over n\)
