Solution : Since, total number of students = 100 and number of boys = 70 \(\therefore\) number of girls = (100 – 70) = 30 Now, the total marks of 100 students = 100*72 = 7200 And total marks of 70 boys = 70*75 = 5250 Total marks of 30 girls = 7250 – 5250 […]
Solution : Given that, for binomial distribution mean, np = 4 and variance, npq = 2. \(\therefore\) q = 1/2, but p + q = 1 \(\implies\) p = 1/2 and n \(\times\) \(1\over 2\) = 4 \(\implies\) n = 8 We know that, P(X = r) = \(^nC_r p^r q^{n-r}\) \(\therefore\) P(X = 1)
Solution : Median of new set remains the same as of the original set. 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 Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\)
Solution : Given that, mean = 4 \(\implies\) np = 4 And Variance = 2 \(\implies\) npq = 2 \(\implies\) 4q = 2 \(\implies\) q = \(1\over 2\) \(\therefore\) p = 1 – q = 1 – \(1\over 2\) = \(1\over 2\) Also, n = 8 Probability of 2 successes = P(X = 2) =
Solution : In the 2n observations, half of them equals a and remaining half equal -a. Then, the mean of total 2n observations is equal to 0. \(\therefore\) SD = \(\sqrt{\sum(x – \bar{x})^2\over N}\) \(\implies\) 4 = \(\sum{x^2}\over 2n\) \(\implies\) 4 = \(2na^2\over 2n\) \(\implies\) \(a^2\) = 4 \(\therefore\) a = 2 Similar Questions The
Solution : Now, P(X > 1.5) = P(2) + P(3) + …… \(\infty\) = 1 – [P(0) + P(1)] = 1 – \((e^{-2} + {e^{-2}(2)\over 1})\) = 1 – \(3\over e^2\) Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1)
A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to Read More »
Solution : Given that, mean = 21 and median = 22 Using the relation, Mode = 3 median – 2 mean \(\implies\) Mode = 3(22) – 2(21) = 66 – 42 = 24 Similar Questions The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X
Question : Let \(x_1\), \(x_2\), ….. , \(x_n\), be n observations such that \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80. Then, a possible value of among the following is (a) 12 (b) 9 (c) 18 (d) 15 Solution : Given \(\sum{x_i}^2\) = 400 and \(\sum{x_i}\) = 80 \(\because\) \(\sigma^2\) \(\ge\) 0 \(\therefore\) \(\sum{x_i}^2\over n\) –
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
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 =
