This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
Here's how to solve this probability density function problem. Part (a): Show that k = (3)/(4) and compute the exact mean E(X) and variance Var(X). Step 1: Show that k = (3)/(4). For f(x) to be a valid probability density function, the integral over its entire domain must equal 1. _-^ f(x)\,dx = 1 Since f(x) = kx(2-x) for 0 x 2 and 0 otherwise, we integrate from 0 to 2: _0^2 kx(2-x)\,dx = 1 k _0^2 (2x-x^2)\,dx = 1 k [ x^2 - (x^3)/(3) ]_0^2 = 1 k ( (2^2 - (2^3)/(3)) - (0^2 - (0^3)/(3)) ) = 1 k ( 4 - (8)/(3) ) = 1 k ( (12)/(3) - (8)/(3) ) = 1 k ( (4)/(3) ) = 1 k = (3)/(4) Thus, k = (3)/(4). Step 2: Compute the exact mean E(X). The mean E(X) is given by _-^ x f(x)\,dx. E(X) = _0^2 x · (3)/(4) x(2-x)\,dx E(X) = (3)/(4) _0^2 (2x^2 - x^3)\,dx E(X) = (3)/(4) [ (2x^3)/(3) - (x^4)/(4) ]_0^2 E(X) = (3)/(4) ( ( (2(2)^3)/(3) - ((2)^4)/(4) ) - (0) ) E(X) = (3)/(4) ( (16)/(3) - (16)/(4) ) E(X) = (3)/(4) ( (16)/(3) - 4 ) E(X) = (3)/(4) ( (16-12)/(3) ) = (3)/(4) ( (4)/(3) ) = 1 The exact mean is E(X) = 1. Step 3: Compute the exact variance Var(X). First, we compute E(X^2) = _-^ x^2 f(x)\,dx. E(X^2) = _0^2 x^2 · (3)/(4) x(2-x)\,dx E(X^2) = (3)/(4) _0^2 (2x^3 - x^4)\,dx E(X^2) = (3)/(4) [ (2x^4)/(4) - (x^5)/(5) ]_0^2 E(X^2) = (3)/(4) [ (x^4)/(2) - (x^5)/(5) ]_0^2 E(X^2) = (3)/(4) ( ( (2^4)/(2) - (2^5)/(5) ) - (0) ) E(X^2) = (3)/(4) ( (16)/(2) - (32)/(5) ) E(X^2) = (3)/(4) ( 8 - (32)/(5) ) E(X^2) = (3)/(4) ( (40-32)/(5) ) = (3)/(4) ( (8)/(5) ) = (6)/(5) Now, we use the formula Var(X) = E(X^2) - (E(X))^2. Var(X) = (6)/(5) - (1)^2 Var(X) = (6)/(5) - 1 = (1)/(5) The exact variance is Var(X) = (1)/(5). Part (b): Find the cumulative distribution function F(x) and evaluate the probability P(0.5 X 1.5). Step 1: Find the cumulative distribution function F(x). The CDF F(x) is defined as F(x) = _-^x f(t)\,dt. For x < 0: F(x) = _-^x 0\,dt = 0 For 0 x 2: F(x) = _0^x (3)/(4)t(2-t)\,dt F(x) = (3)/(4) _0^x (2t-t^2)\,dt F(x) = (3)/(4) [ t^2 - (t^3)/(3) ]_0^x F(x) = (3)/(4) ( x^2 - (x^3)/(3) ) = (3x^2)/(4) - (x^3)/(4) For x > 2: F(x) = _0^2 (3)/(4)t(2-t)\,dt = 1 Combining these, the cumulative distribution function is: F(x) = 0, & x < 0 \\ (3x^2)/(4) - (x^3)/(4), & 0 x 2 \\ 1, & x > 2 Step 2: Evaluate the probability P(0.5 X 1.5). We can calculate this probability using the CDF: P(0.5 X 1.5) = F(1.5) - F(0.5). Using F(x) = (3x^2)/(4) - (x^3)/(4): F(1.5) = (3(1.5)^2)/(4) - ((1.5)^3)/(4) = (3(9)/(4))4 - (27)/(8)4 = (27)/(16) - (27)/(32) = (54)/(32) - (27)/(32) = (27)/(32) F(0.5) = (3(0.5)^2)/(4) - ((0.5)^3)/(4) = (3(1)/(4))4 - (1)/(8)4 = (3)/(16) - (1)/(32) = (6)/(32) - (1)/(32) = (5)/(32) P(0.5 X 1.5) = F(1.5) - F(0.5) = (27)/(32) - (5)/(32) = (22)/(32) = (11)/(16) The probability is P(0.5 X 1.5) = (11)/(16). That's 2 down. 3 left today — send the next one.