HW3

1. Suppose a continuous random variable X has the following probability density function (a) Find the value of k that makes fX(x) a valid probability density function. (Recall a property that a PDF must have) f X (x)  0 for all x, we have x(1-x)  0 for all 0  x  1 => k  0 (1) ∫ − ∞ ∞ fX ( x ) dx = 1 => ∫ − ∞ 0 0 dx + ∫ 0 1 kx ( 1 − x ) dx + ∫ 0 ❑ 0 dx = 1 => ∫ 0 1 kx ( 1 − x ) dx = 1 => ∫ 0 1 kxdx - ∫ 0 1 k x 2 dx = 1 => kx 2 /2| 1 0 – kx 3 /3| 1 0 = 1 => k ( 1/2 - 1/3 ) = 1 => 1/6 k = 1 => k = 6 (2) (1) and (2) => k = 6 (b) Give the CDF, F X (x). F X (x) = P (X  x) For x < 0: F X (x) = 0 For x > 1: F X (x) = 1 For 0  x  1: F X (x) = ∫ 0 x 6 x ( 1 − x ) dx = 3x 2 | x 0 – 2x 3 | x 0 = 3x 2 – 2x 3 F X (x) = { 0 ,x < 0 3 x 2 – 2 x 3 , 0 x 1 1 ,x > 1

(c) Find P(0.5 ≤ X ≤ 1) using f X (x). P(0.5 ≤ X ≤ 1) = ∫ 0.5 1 6 x ( 1 − x ) dx = 0.5 (d) Find P(0 ≤ X ≤ 0.75) using F X (x). P(0 ≤ X ≤ 0.75) = F X (0.75) – F X (0) = 3 ( 0.75 ) 2 – 2 ( 0.75 ) 3 – 0 = 0.84375 (e) Find E(X). E(X) = ∫ 0 1 6 x 2 ( 1 − x ) dx = ∫ 0 1 6 x 2 dx - ∫ 0 1 6 x 3 dx = 2x 3 1 0 - 3/2 x 4 1 0 = 0.5 (f) Find Var(X). E(X 2 ) = ∫ 0 1 6 x 3 ( 1 − x ) dx = ∫ 0 1 6 x 3 dx - ∫ 0 1 6 x 4 dx = 3/2x 4 1 0 - 6/5 x 5 1 0 = 3/10 Var(X) = E(X 2 ) – E(X) 2 = 3/10 – (1/2) 2 = 0.05 2. Suppose a continuous random variable X has the following probability density function: (a) Write down the cumulative distribution function (CDF) in functional form. (Make sure to cover all cases). F X (t) = P (X  t) For t < -1: F X (t) = 0 For t > 1: F X (t) = 1 For -1  t  1: F X (x) = ∫ − 1 t 3 2 x 2 dx = 1 2 x 3 | t − 1 = 1 2 (t 3 +1) F X (t) = { 0 ,t ← 1 1 2 ( t 3 + 1 ) , − 1 t 1 1 ,t > 1

(b) Use your CDF from part (a) to find P(|X – 1/2| < 1/4). (Write an expression in terms of F X first, then solve). P(|X – 1/2| < 1/4) = P ( -1/4 < X-1/2 < 1/4) = P ( ¼ < X < ¾) = F X (3/4) – F X (1/4) = 1 2 [ ( 3 4 3 + 1 ) − ( 1 4 3 + 1 ) ] = 0.203125 (c) Find the value t such that P(X ≤ t) = 0.80. = F X (t) = 0.80 = 1 2 ( t 3 + 1 ) => t 3 + 1 = 1.6 => t = 0.8434 3. A supplier of kerosene has a 150-gallon tank that is filled at the beginning of each week. His weekly demand shows a relative frequency behavior that increases steadily up to 100 gallons and then levels off between 100 and 150 gallons. If X denotes weekly demand in hundreds of gallons, the relative frequency of demand can be modeled by (a) Find F X (x). (Remember to cover all cases) F X (x) = P (X  x) For x < 0: F X (x) = 0 For x > 1.5: F X (x) = 1 For 0  x  1 : F X (x) = ∫ 0 x xdx = x 2 2 x 0 = x 2 2 For 1 < x  1.5: F X (x) = ∫ 0 1 xdx + ∫ 1 x dx = 1 2 + x – 1 = x - 1 2 F X (x) = { 0 , x < 0 x 2 2 , 0 x 1 x − 1 2 , 1 < x 1.5 1 , x > 1.5