Statistics Questions

Solution : Since variance is independent of change of origin. Therefore, variance of observations 101, 102, …. , 200 is same as variance of 151, 152, ….. 250. \(\therefore\)  \(V_A\) = \(V_B\) \(\implies\)   \(V_A\over V_B\) = 1 Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and …

Suppose a population A has 100 observation 101, 102, ….. , 200 and another population B has 100 observations 151, 152, …. , 250. If \(V_A\) and \(V_B\) represent the variance of the two populations respectively, then \(V_A\over V_B\) is Read More »

Solution : Let the number of boys and girls be x and y, respectively \(\therefore\)   52x + 42y = 50(x + y) \(\implies\)  52x + 42y = 50x + 50y \(\implies\)  2x = 8y \(\implies\)  x = 4y \(\therefore\)  Total number of students in the class = x + y = 4y + y = …

The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys and girls combined is 50. The percentage of boys in the class is Read More »

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 …

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then, find the values of a and b? Read More »

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 …

A scientist is weighing each of 30 fishes. Their mean weight worked out is 30 g and a standard deviation of 2 g. Later, it was found that the measuring scale was misaligned and always under reported every fish weight by 2 g. The correct mean and standard deviation in gram of fishes are respectively. Read More »

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 …

If the mean deviation about the median of numbers a, 2a, …. , 50a is 50, then |a| is equal to Read More »

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 …

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 ? Read More »

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 \(\implies\) …

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be Read More »

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 = …

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than Read More »