# Statistics Questions

## 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 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 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) = …

## 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 …

## 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) …

## 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 …

## 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 $$\therefore$$  $$\sum{x_i}^2\over n$$ – …

## 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 …

## 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 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?

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 …