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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 »

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that three apply for the same house is

Solution : \(\therefore\)  Total number of cases = \(3^3\) Now, favourable cases = 3(as either all has applied for house 1 or 2 or 3) \(\therefore\) Required Probability = \(3\over 3^3\) = \(1\over 9\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to …

Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that three apply for the same house is 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 »

At a telephone enquiry system, the number of phone calls regarding relevant enquiry follow. Poisson distribution with an average of 5 phone calls during 10 min time intervals. The probability that there is almost one phone call during a 10 min time period is

Solution : Required Probability = P(X = 0) + P(X = 1) = \(e^{-5}\over 0!\).\(5^0\) + \(e^{-5}\over 1!\).\(5^1\) = \(e^{-5}\) + 5\(e^{-5}\) = \(6\over {e^5}\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence from match to match. The probability that in a match series India’s …

At a telephone enquiry system, the number of phone calls regarding relevant enquiry follow. Poisson distribution with an average of 5 phone calls during 10 min time intervals. The probability that there is almost one phone call during a 10 min time period 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 »

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is

Solution : Let the events, A = Ist aeroplane hit the target B = 2nd aeroplane hit the target And their corresponding probabilities are P(A) = 0.3 and P(B) = 0.2 \(\implies\) P(A’) = 0.7 and P(B’) = 0.8 \(\therefore\)  Required Probability = P(A’)P(B) + P(A’)P(B’)P(A’)P(B) + ……. = (0.7)(0.2) + (0.7)(0.8)(0.7)(0.2) + ……. = …

Two aeroplanes I and II bomb a target in succession. The probabilities of I and II scoring a hit correctly are 0.3 and 0.2, respectively. The second plane will bomb only if the first misses the target. The probability that the target is hit by the second plane, is Read More »

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is

Solution : Probability of getting score 9 in a single throw = \(4\over 36\) = \(1\over 9\) Probability of getting score 9 exactly in double throw = \(^3C_2\) \(\times\) \(({1\over 9})^2\) \(\times\) \(8\over 9\) = \(8\over 243\) Similar Questions The probability of India winning a test match against the west indies is 1/2 assuming independence …

A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is Read More »

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 …

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 »

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