Question 7.4: A quality index summarizes different features of a product b......
A quality index summarizes different features of a product by means of a score. Different experts may assign different quality scores depending on their experience with the product. Let X be the quality index for a tablet. Suppose the respective probability density function is given as follows:
f\left(x\right) =\left\{\begin{matrix} cx(2 − x) if 0 \leq x \leq 2 \\ 0 elsewhere. \end{matrix} \right.
(a) Determine c such that f (x) is a proper PDF.
(b) Determine the cumulative distribution function.
(c) Calculate the expectation and variance of X.
(d) Use Tschebyschev’s inequality to determine the probability that X does not deviate more than 0.5 from its expectation.
(a) Two conditions need to be satisfied for f (x) to be a proper PDF:
(i) \int_{0}^{2}{f\left(x\right)dx} =1:
\int_{0}^{2}{f\left(x\right)dx}= \int_{0}^{2}{c \cdot x \left(2-x\right)dx}=c \int_{0}^{2}{ x \left(2-x\right)dx}
= c \int_{0}^{2}{\left(2x-x^{2}\right)dx}= c \left[x^{2}- \frac{1}{3}x^{3}\right] ^{2}_{0}
= c \left[4- \frac{8}{3}-\left(0-0\right)\right] = c \cdot \frac{4}{3}\overset{!}{=} 1
\Longrightarrow c= \frac{3}{4}.
(ii) f (x) ≥ 0:
f (x) = \frac{3}{4}x (2 − x) ≥ 0 ∀ x ∈ [0, 2].
(b) We calculate
F(x) = P(X ≤ x) = \int_{0}^{x}{f\left(t\right)dt} = \int_{0}^{x}{\frac{3}{4} t \left(2- t\right)dt}
=\frac{3}{4}\int_{0}^{x}{\left(2t- t^{2}\right)dt}= \frac{3}{4}\left[t^{2}- \frac{1}{3}t^{3}\right] ^{x}_{0}
= \frac{3}{4}\left[x^{2}- \frac{1}{3}x^{3}-0\right] = \frac{3}{4}x^{2} \left(1-\frac{1}{3}x\right)
and therefore
F(x) = \left\{\begin{matrix} 0\quad\quad \quad\quad if x \lt 0 \\ \frac{3}{4}x^{2} \left(1-\frac{1}{3}x\right)\quad if 0 \leq x \leq 2\\ 1\quad\quad \quad\quad if 2 \lt x.\end{matrix} \right.
(c) The expectation is
E(X) = \int_{0}^{2}{x f\left(x\right)dx}= \frac{3}{4}\int_{0}^{2}{\left(2x^{2} – x^{3}\right)dx}
= \frac{3}{4}\left[\frac{2}{3}x^{3}-\frac{1}{4}x^{4}\right] ^{2}_{0}= \frac{3}{4}\left[\frac{2}{3} \cdot 8 – \frac{1}{4} \cdot 16 -0\right]
= \frac{3}{4}\left[\frac{16}{3}-\frac{12}{3}\right] = \frac{3}{4}\cdot \frac{4}{3}=1.
Using Var(X) = E(X²) − (E(X))², we calculate the variance as
E(X²) = \int_{0}^{2}{x^{2} f\left(x\right)dx}= \frac{3}{4}\int_{0}^{2}{\left(2x^{3} – x^{4}\right)dx}
= \frac{3}{4}\left[\frac{2}{4}x^{4}-\frac{1}{5}x^{5}\right] ^{2}_{0}= \frac{3}{4}\left[\frac{2}{4} \cdot 16 – \frac{1}{5} \cdot 32-0\right]
= 6- \frac{3\cdot 32}{4\cdot 5} = 6- \frac{3\cdot 8}{5}=\frac{6}{5}
Var(X) = \frac{6}{5}-1^{2} =\frac{1}{5}.
(d)
P\left(\left|X-\mu \right| \leq 0.5\right)\geq 1-\frac{\sigma ^{2}}{c^{2}} =1-\frac{\left(\frac{1}{5} \right) }{\left(0.5\right)^{2} } =1-0.8=0.2.