Statistics Questions

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls?

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 …

In a class of 100 students. there are 70 boys whose average marks in a subject are 75. If the average marks of the complete class is 72, then what is the average of the girls? Read More »

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is

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) …

The mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, then P(X = 1) is Read More »

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is

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 \(\sum{x_i}\) …

The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observation of the set is increased by 2, then the median of the new is Read More »

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is

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) = …

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 success is Read More »

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to

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 …

In a series of 2n observations, half of them equals a and remaining half equal -a. If the standard deviation of the observation is 2, then |a| equal to Read More »

A random variable X has poisson distribution with mean 2. Then, P(X > 1.5) equal to

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 »

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately

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 …

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately Read More »

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

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 \(\therefore\)  \(\sum{x_i}^2\over n\) – …

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 Read More »

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

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 »

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

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 »