If a r . v . X has p . d . f . , f x = c x , for 1 < x < 3 , c > 0 , Find c , E X and Var X .

Given that: A random variable X has p.d.f. f ( x ) = c x , for 1 < x < 3 We know that, in probability density function, ∫ a b f ( x ) d x = 1 ⇒ ∫ 1 3 c x d x = 1 ⇒ c ∫ 1 3 1 x d x = 1 ⇒ c log x 1 3 = 1 ⇒ c log 3 - log 1 = 1 ⇒ c log 3 - 0 = 1 ⇒ c = 1 log 3 Mean (or) Expected value of random variable X is E ( X ) = ∫ 1 3 x f ( x ) d x = ∫ 1 3 x · c x d x = c ∫ 1 3 d x = c x 1 3 = 1 log 3 3 - 1 ∴ E X = = 2 log 3 Now, consider, ∫ 1 3 x 2 f ( x ) d x = ∫ 1 3 x 2 · c x d x = c ∫ 1 3 x d x = c x 2 2 1 3 = 1 2 log 3 9 - 1 = 4 log 3 Now, Var X = ∫ 1 3 x 2 f ( x ) d x - [ E ( X ) ] 2 = 4 log 3 - 2 log 3 2 = 4 log 3 - 4 log 3 2 = 4 log 3 - 1 log 3 2 Hence, c = 1 log 3 , E x = 2 log 3 and V a r x = 4 log 3 - 1 log 3 2 .

If a r . v . X has p . d . f . , f x = c x , for 1 < x < 3 , c > 0 , Find c , E X and Var X .

Given that: A random variable X has p.d.f. f ( x ) = c x , for 1 < x < 3 We know that, in probability density function, ∫ a b f ( x ) d x = 1 ⇒ ∫ 1 3 c x d x = 1 ⇒ c ∫ 1 3 1 x d x = 1 ⇒ c log x 1 3 = 1 ⇒ c log 3 - log 1 = 1 ⇒ c log 3 - 0 = 1 ⇒ c = 1 log 3 Mean (or) Expected value of random variable X is E ( X ) = ∫ 1 3 x f ( x ) d x = ∫ 1 3 x · c x d x = c ∫ 1 3 d x = c x 1 3 = 1 log 3 3 - 1 ∴ E X = = 2 log 3 Now, consider, ∫ 1 3 x 2 f ( x ) d x = ∫ 1 3 x 2 · c x d x = c ∫ 1 3 x d x = c x 2 2 1 3 = 1 2 log 3 9 - 1 = 4 log 3 Now, Var X = ∫ 1 3 x 2 f ( x ) d x - [ E ( X ) ] 2 = 4 log 3 - 2 log 3 2 = 4 log 3 - 4 log 3 2 = 4 log 3 - 1 log 3 2 Hence, c = 1 log 3 , E x = 2 log 3 and V a r x = 4 log 3 - 1 log 3 2 .

From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs are drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Out of 30 bulbs, 6 are defective. ⇒ Number of non-defective bulbs = 30 - 6 = 24 Probability of drawing a defective bulb = 6 30 = 1 5 Probability of drawing a non-defective bulb = 24 30 = 4 5 4 bulbs are drawn from the lot with replacement. Let X be the random variable that denotes the number of defective bulbs in the selected bulbs. ∴ P X = 0 = P ( 4 non-defective and 0 defective) = C 0 4 × 4 5 4 × 1 5 0 = 256 625 P X = 1 = P ( 3 non-defective and 1 defective) = C 1 4 × 1 5 × 4 5 3 = 256 625 P X = 2 = P ( 2 non-defective and 2 defective) = C 2 4 × 1 5 2 × 4 5 2 = 96 625 P X = 3 = P ( 1 non-defective and 3 defective) = C 3 4 × 1 5 3 × 4 5 = 16 625 P X = 4 = P ( 0 non-defective and 4 defective) = C 4 4 × 1 5 4 × 4 5 0 = 1 625 Therefore, the required probability distribution is as follows: X 0 1 2 3 4 P X 256 625 256 625 96 625 16 625 1 625

A random variable X has the following probability distribution: X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Find k .

Given that, X 0 1 2 3 4 5 6 7 P X 0 k 2 k 2 k 3 k k 2 2 k 2 7 k 2 + k Since the sum of all the probability in a probability distribution is always unity. ∴ P ( X = 0 ) + P ( x = 1 ) + . . . . . . + P ( x = 7 ) = 1 = 0 + k + 2 k + 2 k + 3 k + k 2 + 2 k 2 + 7 k 2 + k = 1 = 10 k 2 + 9 k - 1 = 0 ⇒ ( 10 k - 1 ) ( k + 1 ) = 0 ⇒ k ≠ - 1 ∵ k ≥ 0 ⇒ 10 k - 1 = 0 ⇒ k = 1 10