For the following probability density function (p.d.f) of X , f x = x 2 18 , − 3 < x < 3 0 , otherwise , find P ( X < 1 ) .

Given, f x = x 2 18 , − 3 < x < 3 0 , otherwise is a p.d.f Now, P X < 1 = P - 1 < X < 1 = ∫ − 1 1 f ( x ) d x ⇒ P X < 1 = ∫ − 1 1 x 2 18 d x ⇒ P X < 1 = 1 18 x 3 3 - 1 1 ⇒ P X < 1 = 1 18 1 3 3 - - 1 3 3 ⇒ P X < 1 = 1 18 1 3 - - 1 3 ⇒ P X < 1 = 1 18 1 3 + 1 3 ⇒ P X < 1 = 1 18 2 3 ⇒ P X < 1 = 1 27 = 0 . 03704

For the following probability density function (p.d.f) of X , f x = x 2 18 , − 3 < x < 3 0 , otherwise , find P ( X < 1 ) .

Given, f x = x 2 18 , − 3 < x < 3 0 , otherwise is a p.d.f Now, P X < 1 = P - 1 < X < 1 = ∫ − 1 1 f ( x ) d x ⇒ P X < 1 = ∫ − 1 1 x 2 18 d x ⇒ P X < 1 = 1 18 x 3 3 - 1 1 ⇒ P X < 1 = 1 18 1 3 3 - - 1 3 3 ⇒ P X < 1 = 1 18 1 3 - - 1 3 ⇒ P X < 1 = 1 18 1 3 + 1 3 ⇒ P X < 1 = 1 18 2 3 ⇒ P X < 1 = 1 27 = 0 . 03704

Find k if the function f x is defined by f x = k x 1 - x , for 0 < x < 1 = 0 , otherwise , is the probability density function (p.d.f.) of a random variable (r.v.) X . Also find P x < 1 2 .

Given that the function f x = k x 1 - x , for 0 < x < 1 = 0 , otherwise , is the p.d.f. of X . ⇒ ∫ - ∞ ∞ f ( x ) d x = 1 ⇒ ∫ - ∞ 0 f ( x ) d x + ∫ 0 1 f ( x ) d x + ∫ 1 ∞ f ( x ) d x = 1 ⇒ 0 + ∫ 0 1 k x ( 1 - x ) d x + 0 = 1 ⇒ k ∫ 0 1 x - x 2 d x = 1 ⇒ k x 2 2 - x 3 3 0 1 = 1 ⇒ k 1 2 - 1 3 - 0 = 1 ⇒ k 6 = 1 Hence, k = 6 Now, we need to find: P x < 1 2 We know that P x < 1 2 = ∫ - ∞ 1 2 f ( x ) d x = ∫ - ∞ 0 f ( x ) d x + ∫ 0 1 2 f ( x ) d x = 0 + ∫ 0 1 2 k x ( 1 - x ) d x = k ∫ 0 1 2 x - x 2 d x = k x 2 2 - x 3 3 0 1 2 = k 1 8 - 1 24 - 0 = k 2 24 = 6 × 2 24 Hence, P x < 1 2 = 1 2 .

Obtain the probability distribution of the number of sixes in two tosses of a fair die.

When two dice are tossed then the sample space, n S = 6 × 6 = 36 . . . 1 Let X denote the number of sixes in two tosses of dice. So, X = 0 , 1 , 2 Case 1 . When X = 0 {i.e. no six came.} Then, x = ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 1 , 4 ) , ( 1 , 5 ) , 2 , 1 , 2 , 2 , 2 , 3 , 2 , 4 , 2 , 5 , 3 , 1 , ( 3 , 2 ) , ( 3 , 3 ) , ( 3 , 4 ) , ( 3 , 5 ) , ( 4 , 1 ) , ( 4 , 2 ) , ( 4 , 3 ) , ( 4 , 4 ) , ( 4 , 5 ) , ( 5 , 1 ) , ( 5 , 2 ) , ( 5 , 3 ) , ( 5 , 4 ) , ( 5 , 5 ) Thus, n x = 25 So, P X = 0 = n x n S = 25 36 . . . 2 Case 2 . When X = 1 {i.e. only one six came.} So, P ( X = 1 ) = n ( x ) n ( S ) Then, x = 1 , 6 , 2 , 6 , 3 , 6 , 4 , 6 , 5 , 6 , 6 , 1 , 6 , 2 , 6 , 3 , 6 , 4 , 6 , 5 i.e. n x = 10 Thus, P ( X = 1 ) = 10 36 = 5 18 . . . 3 Case 3 . When X = 2 {i.e. both dice shows six.} So, x = { ( 6 , 6 ) } ⇒ n x = 1 Thus, P ( X = 2 ) = n ( x ) n ( s ) = 1 36 . . . 4 From all cases probability distribution of number of sixes is X 0 1 2 P X 25 36 10 36 1 36

For the following probability density function (p.d.f) of X , f x = x 2 18 , − 3 < x < 3 0 , otherwise , find P ( X < 1 ) .

Given, f x = x 2 18 , − 3 < x < 3 0 , otherwise is a p.d.f Thus, P X < 1 = ∫ − 3 1 f ( x ) d x ⇒ P X < 1 = ∫ − 3 1 x 2 18 d x ⇒ P X < 1 = 1 18 x 3 3 - 3 1 ⇒ P X < 1 = 1 18 1 3 3 - - 3 3 3 ⇒ P X < 1 = 1 18 1 3 - - 27 3 ⇒ P X < 1 = 1 18 1 3 + 9 ⇒ P X < 1 = 0 . 5185

The probability distribution of a random variable X , the number of defects per 10 meters of a fabric is given by: x 0 1 2 3 4 P X = x 0 . 45 0 . 35 0 . 15 0 . 03 0 . 02 Find the variance of X .

Solution: x i p i p i x i p i x i 2 0 0 . 45 0 0 1 0 . 35 0 . 35 0 . 35 2 0 . 15 0 . 30 0 . 60 3 0 . 03 0 . 09 0 . 27 4 0 . 02 0 . 08 0 . 32 Total 0 . 82 1 . 54 From the table, ∑ p i x i = 0.82 and ∑ p i x i 2 = 1.54 So, Var X = ∑ p i x i 2 − ∑ p i x i 2 = 1.54 − ( 0.82 ) 2 = 1.54 − 0.6724 = 0.8676 Thus, Var X = 0 . 8676