Answer to Question #97238 in Statistics and Probability for Juliet Beglaryan

Question

1. The thickness x of a protective coating applied to a conductor designed to work in corrosive conditions follows a uniform distribution over the interval [20, 40] microns. a) What is the probability of the thickness of the protective coating applied to the conductor to be less than 35 microns?b) Find the probability of the thickness of the protective coating applied to the conductor to be between 23 and 32 microns.c) Find the mean, standard deviation of the thickness of the protective coating. Find also the probability that the coating is less than 35 microns thick.

Answer 1

We know that $X \sim U(20,40)$ with probability density

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

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

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

a) We asked to find $P(X < 35)$

$P(X < 35) = {F_X}(35) = \frac{{35 - 20}}{{40 - 20}} = \frac{3}{4}$

b) We asked to find $P(23 < X < 32)$

$P(23 < X < 32) = {F_X}(32) - {F_X}(23) = \frac{{32 - 20}}{{40 - 20}} - \frac{{23 - 20}}{{40 - 20}} = \frac{9}{{20}}$

c) We asked to find the expected value and standard deviation. Let's calculate the expected value using it's definition

$\mathbb{E}X = \int\limits_{ - \infty }^{ + \infty } {x{p_X}(x)dx} = \int\limits_a^b {\frac{x}{{b - a}}dx} = \left. {\frac{{{x^2}}}{{2(b - a)}}} \right|_a^b = \frac{{{b^2} - {a^2}}}{{2(b - a)}} = \frac{{a + b}}{2}$

Thus in our case

$\mathbb{E}X = \frac{{a + b}}{2} = \frac{{20 + 40}}{2} = 30$

To find the standart deviatin we need to find the variance at the first. To find the variance let's find the second moment

$\mathbb{E}({X^2}) = \int\limits_{ - \infty }^{ + \infty } {{x^2}{p_X}(x)dx} = \int\limits_a^b {\frac{{{x^2}}}{{b - a}}dx} = \left. {\frac{{{x^3}}}{{3(b - a)}}} \right|_a^b = \frac{{{a^2} + ab + {b^2}}}{3}$

Now we can calculate the variance

$\operatorname{Var} (X) = \mathbb{E}({X^2}) - {\mathbb{E}^2}(X) = \frac{{{a^2} + ab + {b^2}}}{3} - \frac{{{{(a + b)}^2}}}{{{2^2}}} = \frac{1}{{12}}{(b - a)^2}$

And the standard deviations is

$\sigma (X) = \sqrt {\frac{1}{{12}}{{(a - b)}^2}} = \frac{{b - a}}{{2\sqrt 3 }}$

In our case

$\sigma (X) = \frac{{40 - 20}}{{2\sqrt 3 }} = \frac{{10}}{{\sqrt 3 }}$