Question

2. Let X denote the amount of space occupied by an article placed in a 1-ft packing container. The pdf of X is f(x)=90x8 (1 – x) for 0<x< 1 and zero otherwise a) What is the CDF of X b) What is P(X < 0.5) ? c) What is P(0.25 <X < 0.5? And P(0.25 < X <0.5) ? d) Compute E(X)and V(X) e) What is the probability that X is more than one standard deviation from its mean value?

Answer 1

$\small \text{Given that the pdf of } X \text{ is }, \\ f(x) = \begin{cases} 90x^8(1-x); & 0 \le x \le 1 \\ 0; & \text{otherwise} \end{cases}$

a) The CDF, F(x) of X is obtained as, 0; <0 F(x) = { so f(u)du; Or 1 1; r>1 1900(1 - 1)du Now, Virtuyat = //i9 on / (au - 1) = 90 9 101 = 90 - 9 10 9 2-101 = 90 9 10 = 79 (10 - 9.7) 10; T<0 So, F(x) = 3 79 (10 - 9.7); Or 1 11: I>1

$\small \begin{align*} b) \ P(X \le 0.5) &= F(0.5) \\ &= 0.5^9(10 - 9*0.5) \\ &= \frac{1}{2^9}\left(10 - \frac{9}{2} \right ) \\ &= \frac{11}{2^{10}} \\ &\approx 0.01074 \end{align*}$

$\small c) \text{ We observe that the CDF defined by } F(x) \text{ is an absolutely continuous function. Thus for all } x, \\ \lim_{y \to x^{-}}F(y) = F(x^-) = F(x) = F(x^+) = \lim_{y\to x^+}F(y) \\ \text{Hence, for all } a < b \in \mathbb{R}, \\ P(a < X \le b) = F(b) - F(a) = F(b) - F(a^-) = P(a \le X \le b)\\ \begin{align*} \text{So, } P(0.25 < X \le 0.5) = P(0.25 \le X \le 0.5) &= F(0.5) - F(0.25) \\ &= \frac{1}{2^9} \left(10 - \frac{9}{2} \right) - \frac{1}{4^9}\left(10 - \frac{9}{4} \right ) \\ &= \frac{11}{2^{10}} - \frac{31}{4^{10}} \\ &\approx 0.01071 \end{align*}$

$\small d) \text{ Consider, } E(X^r), \ \ r \in \mathbb{R} - \{-9, -10\} \\ \begin{align*} E(X^r) &= \int_0^1 x^rf(x)dx \\ &= 90\int_0^1x^{8+r}(1-x) dx \\ &= 90\int_0^1(x^{8+r} - x^{9+r})dx \\ &= 90 \left[\frac{x^{9+r}}{9+r} - \frac{x^{10+r}}{10+r} \right ]_{x=0}^{x=1} \\ &= 90\left[\frac{1}{9+r} - \frac{1}{10+r} \right ] \\ &= \frac{90}{(9+r)(10+r)} \\ \text{So, } E(X) &= \frac{90}{(9+1)(10+1)} \\ &= \frac{9}{11} \\ \text{Also, } E(X^2) &= \frac{90}{(9+2)(10+2)} \\ &= \frac{15}{22} \\ \text{Thus, Var}(X) &= E(X^2) - (E(X))^2 \\ &= \frac{15}{22} - \left( \frac{9}{11}\right)^2 \\ &= \frac{3}{242} \end{align*}$

$\small d) \text{ We need, } P\left(|X - E(X)| > \sqrt{\text{Var}(X)}\right) = 1 - P\left(|X - E(X)| \le \sqrt{\text{Var}(X)}\right)\\ \begin{align*} \text{Now, } P\left(|X - E(X)| \le \sqrt{\text{Var}(X)}\right) &= P\left(E(X) - \sqrt{\text{Var}(X)} \le X \le E(X) + \sqrt{\text{Var}(X)} \right) \\ &= P\left( \frac{9}{11} - \sqrt{\frac{3}{242}} \le X \le \frac{9}{11} + \sqrt{\frac{3}{242}} \right ) \\ &= F\left(\frac{9}{11} + \sqrt{\frac{3}{242}} \right ) - F\left( \frac{9}{11} - \sqrt{\frac{3}{242}}\right ) \\ &\approx F(0.92952) - F(0.70684) \\ &\approx 0.68633 \\ \text{Then, } P\left(|X - E(X)| > \sqrt{\text{Var}(X)}\right) &= 1 - 0.68633 \\ &= 0.31367 \end{align*}$