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**Probability Distribution**

One Mark Questions Probability Distribution PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

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If f(x) = kx2, 0 < x < , elsewhere is a probability density function then the value of k is (a) 1/3 (b) 1/6 (c) 1/9 (d) 1/12 If f(x) = – < x < is a p.d.f of a continuous random variable X, then the value of A is (a) (b) 8 (c) (d) 1

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A random variable X has the following probability distribution X P(X = x) ¼ 2a 3a 4a 5a ¼ then P(1 X 4) (a) 10/21 (b) 2/7 (c) 1/ (d) 1/2

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A random variable X has the following probability mass function as follows: X P(X = x) /6 /4 /12 then the value of is (a) (b) 2 (c) (d) 4

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A random variable X has the following p.d.f then the value of is (a) (b) (c) (d) –1 or X 1 2 3 4 5 6 7 P(X = x) k 2k 3k k2 2 k2 7k2+k

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X is a discrete random variable which takes 0, 1, 2 and P(X = 0) = P(X = 1) = then the value of P(X = 2)is (a) (b) (c) (d)

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Given E(X + c) = 8 and E(X – c) = 12 then the value of c is (a) – (b) 4 (c) – (d) 2 X is a random variable taking the value 3, 4 and 7 with probabilities 1/3, ¼ and 4/7 then E(X) is (a) (b) 6 (c) (d) 3

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Variance of the random variable X is 4. Its mean is 2, then E(X2) is (a) (b) 4 (c) (d) 8 2 = 20, 2’ = 276 for a discrete random variable X, then the mean of the random variable X is (a) (b) 5 (c) (d) 1

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Var (4X + 3) is (a) (b) 16 Var(X) (c) (d) 0 In 5 throws of a die, getting 1 or 2 is a success. The mean number of successes is (a) (b) (c) (d)

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The mean of a binomial distribution is 5 and its standard deviation is 2. Then the value of n and p is (a) (b) (c) (d) If in a Poisson distribution P(X = 0) = k, then the variance is (a) log(1/k) (b) log k (c) e (d) 1/k

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If the mean and standard deviation of a binomial distribution are 12 and 2 respectively. Then the value of its parameter p is (a) ½ (b) 1/3 (c) 2/3 (d) ¼ If a random variable X follows Poisson distribution such that E(X2) = 30 then the variance of the distribution is (a) 6 (b) 5 (c) 30 (d) 25

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In 16 throws of a die getting an even number is considered a success. Then the variance of the successes is (a) 4 (b) 6 (c) 2 (d) 256 A box contains 6 red and 4white balls. If 3 balls are drawn at random, the probability of getting 2 white balls is (a) (b) (c) (d)

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For a Poisson distribution with parameter = 0.25 the value of the 2nd moment about the origin is (a) (b) (c) (d) 0.025 If 2 cards are drawn from a well shuffled pack of 52 cards, the probability that they are of the same colours is (a) (b) (c) (d)

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In a Poisson distribution if P(X = 2) = P(X = 3) then the value of its parameter is (a) 6 (b) 2 (c) 3 (d) 0 The distribution function F(X) of a random variable X is (a) a decreasing function (b) an increasing function (c) a constant function (d) increasing first and then decreasing

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If f(x) is a p.d.f of a normal distribution with mean µ then is (a) 1 (b) (c) 0 (d) 0.25 The random variable X follows normal distribution f(x) = then the value of c is (a) (b) (c) (d)

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If f(x) is a p.d.f of a normal variate X and X – N(, 2) then is (a) undefined (b) 1 (c) 0.5 (d) – 0.5 The marks obtained by 400 students in a mathematics test were normally distributed with mean 65. If 120 students got more marks above 85, the number of students securing marks between 45 and 65 is (a) 120 (b) 20 (c) 80 (d) 160

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In a distribution of X, the mean and variance are 1.5 and 0.5, then E(X2) is (a) (b) (c) 2.75 (d) 2.25 The normal probability curve is symmetrical about (a) the x-axis (b) the y-axis (c) the line x = µ (d) the line x =

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In the following probability distribution function the value of ‘a’ is (a) (b) (c) (d) 1 If 3 dice are thrown once, the number of elements in the sample space is (a) 6 (b) 36 (c) 216 (d) none X 1 2 3 4 5 6 P(X) a 2a 3a 4a 5a 6a

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If X is a continuous random variable then P(a < X < b) = (a) P(a X b) (b) P(a < X b) (c) P(a X < b) (d) all the three Two dice are rolled and X be the random variable of getting number of ‘3’s. Then P(X = 2) (a) (b) (c) (d)

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A continuous random variable X has probability density function f(x) then (a) 0 f(x) 1 (b) f(x) 0 (c) f(x) 1 (d) 0 < f(x) < 1 A discrete random variable X has probability mass function p(x) then (a) 0 p(x) 1 (b) p(x) 0 (c) p(x) 1 (d) 0 < p(x) < 1

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A continuous random variable X follows the probability law, f (x) = where A = , then the value of k is (a) A (b) 1/A (c) A2 (d) 0

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If the probability law density function of a random variable is given by f (x) = , then the value of k is (a) 2/3 (b) 3/2 (c) 2 (d) 3

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F (x) = , – < x < is a distribution function of a continuous variable X, then P(x 1) is (a) 3/4 (b) /4 (c) ½ (d) ¼

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If f (x) = is a probability density function of a continuous random variable X, then the value of A is (a) 1 (b) (c) 1/3 (d) log3

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The probability density function of a random variable X is f(x) = then the value of k is (a) / (b) (c) 1/ (d) /

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For the distribution function given by F(x) = , then p.d.f f(x) is (a) (b) x; x > (c) cannot be determined (d) 2x; x<0

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A continuous random variable x has the p.d.f defined by f(x) = , then c is (a) 1 (b) a (c) 1/a (d) – a

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A random variable X has a probability density function f(x) = , then k is (a) 2 (b) (c) (d)

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The mean of the distribution f(x) = is (a) 1/3 (b) (c) – 1/3 (d) – 3 The mean value of p.d.f f(x) = is (a) (b) (c) 1/ (d) 12

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Mean and variance of binomial distribution are (a) nq, npq (b) np, npq (c) np, np (d) np, npq In a binomial distribution if n = 5, P(X = 3) = 2P(X = 2) then (a) p = 2q (b) 2p = q (c) p = q (d) 3p = 2q

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For a Binomial distribution with mean 2 and variance 4/3, p = ? (a) 2/3 (b) 1/3 (c) ¾ (d) 2/3 In a Poisson distribution if P(X = 2) = P(X = 3), then mean = ? (a) 1 (b) (c) 3 (d) 4

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Which of the following is or are correct regarding normal distribution curve? (1) Symmetrical about the line X = mean (2) Mean = median = mode (3) Unimodal (4) Point of inflexion are at X = (a) 1, 2 only (b) 2, 4 only (c) 1, 2, 3 only (d) all

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For a standard normal distribution the mean and variance are (a) , 2 (b) , (c) 0, 1 (d) 1, 1

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The p.d.f of the standard normal variance Z is (z) = (a) (b) (c) (d)

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If X is normally distributed with mean 6 and standard deviation 5 and Z is the corresponding normal variate, then P(0 x 8) = (a) P(– 1.2 < z < .04) (b) P(– 0.12 < z < 0.4) (c) P(– 1.2 < z < 0.4) (d) P(– 0.12 < z < 0.04)

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The probability density function of the normal distribution is f(x) = – < x < , then the mean = (a) (b) (c) – (d) 4

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For the p.d.f of the normal distribution function – < x < , then the mean = (a) 3 (b) 2/ (c) 3/2 (d) 6

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The p.d.f of the normal distribution is – < x < , then the variance 2 = (a) ¼ (b) 1/2 (c) (d) 2

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The p.d.f of the normal distribution is – < x < , then the variance 2 = (a) ½ (b) (c) 3/2 (d) 1/2

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For a normal distribution with mean = 34 and variance 2 = 16, then P(30 < x < 60) = (where X is normal variate and Z is the corresponding standard normal variate) (a) P(–0.25 < z < 0.25) (b) P(0 < z < 1.625) (c) P(0.25 < z < 1.625) (d) P(–0.25 < z < 1.625)

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A random variable X has the following probability mass function, then k is x P(x) k 3k 5k 7k 9k 11k 13k (a) 1/ (b) 1/36 (c) 1/ (d) 1/7

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A discrete random variable takes (a) only a finite number of values (b) all possible values between certain given limits (c) infinite number of values (d) a finite or uncountable number of values

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A continuous random variable takes (a) only a finite number of values (b) all possible values between certain given limits (c) infinite number of values (d) a finite or uncountable number of values

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