Answer to Question #97239 in Statistics and Probability for Juliet Beglaryan

Question

Packages have a nominal net weight of 1 kg. However their actual net weights have a uniform distribution over the interval 980 g to 1030 g. a) Find the probability that the net weight of a package is less than 1 kg. b) Find the probability that the net weight of a package is less than w g, where 980 < w < 1030. c) If the net weights of packages are independent, find the probability that, in a sample of five packages, all five net weights are less than wg and hence find the probability density function of the weight of the heaviest of the packages.

Answer 1

We know that $X \sim U(980,1030)$ with probability density( $a = 980$ and $b = 1030$)

$${p_X}(x)=\begin{cases} 0 & x \notin [a,b] \\ \frac{1}{{b - a}} & x \in [a,b]\end{cases}$$

Let's find the probability density as

$${F_X}(x) = P(X < x) = \int\limits_{ - \infty }^x {{p_X}(\xi )d\xi }$$

At the interval $x \in ( - \infty ,a]$ we get ${F_X}(x) = \int\limits_{ - \infty }^x {0d\xi } = 0$

At the interval $x \in [a,b]$ we get ${F_X}(x) = \int\limits_a^x {\frac{1}{{b - a}}d\xi } = \left. {\frac{\xi }{{b - a}}} \right|_a^x = \frac{{x - a}}{{b - a}}$

At the interval $x \in [b, + \infty )$ we get ${F_X}(x) = \frac{{b - a}}{{b - a}} + \int\limits_b^{ + \infty } {0d\xi } = 1$

Thus we get

$${F_X}(x)=\begin{cases} 0 & x <a \\ \frac{{b - x}}{{b - a}} & a \leqslant x \leqslant b \\ 1 & x > b \end{cases}$$

a) We asked to find $P(X < 1000)$

$$P(X < 1000) = F(1000) = \frac{{1000 - 980}}{{1030 - 980}} = \frac{2}{5}$$

b) We asked to find $P(X < w)$ if $w \in (980,1030)$

$$P(X < w) = F(w) = \frac{{w - 980}}{{1030 - 980}} = \frac{{w - 980}}{{50}}$$

c) Let $Y$ be the number of packages that weight less than $w$ , because the events are independent $Y$ is distributed according to the Binomial distribution $Y \sim B(n,p)$ with $n = 5$ and $p = P(X < w) = \frac{{w - 980}}{{50}}$. The probability mass function for binomial distribtion is

$${p_Y}(k) = C_n^k{p^k}{(1 - p)^{n - k}}$$

where $C_n^k = \frac{{n!}}{{k!(n - k)!}}$ - binomial coefficient. Thus, the probability that all 5 net will weight less than $w$

$$P(Y = 5) = {p_Y}(5) = \frac{{5!}}{{5!0!}}{(\frac{{w - 980}}{{50}})^5}{(1 - \frac{{w - 980}}{{50}})^0} = {(\frac{{w - 980}}{{50}})^5}$$

Now we need to find the probability density for the heaviest package if we know that all 5 packages are weight less than $w$ . Actually, at the previous step we found the conditional probability

$$P(X < w|Y = 5) = {(\frac{{w - 980}}{{50}})^5}$$

Using is we can easily found it

$${f}(w) = \frac{{\partial P(X < w|Y = 5)}}{{\partial w}} = \frac{\partial }{{\partial w}}{(\frac{{w - 980}}{{50}})^5} = \frac{1}{{10}}{(\frac{{w - 980}}{{50}})^4}$$

(of course, for $980 < w < 1030$ )