# Solved:Resistors labeled 100 Omega have true resistances that are between 80 Omega and 120 Omega.

#### ByDr. Raju Chaudhari

Sep 21, 2020

Resistors labeled 100 $\Omega$ have true resistances that are between 80 $\Omega$ and 120 $\Omega$. Let X be the mass of a randomly chosen resistor. The probability density function of X is given by

 \begin{aligned} f(x)=\left\{ \begin{array}{ll} \frac{x-80}{800} & \hbox{80< x< 120;}\\ 0, & \hbox{Otherwise} \end{array} \right. \end{aligned}

a. What proportion of resistors have resistances less than 90 $\Omega$?
b. Find the mean resistance.
c. Find the standard deviation of the resistances.
d. Find the cumulative distribution function of the resistances.

### Solution

a. The proportion of resistors have resistances less than 90 $\Omega$ is

 \begin{aligned} P(X < 90) &= \int_{80}^{90} f(x)\; dx\\ &= \int_{80}^{90} \frac{x-80}{800}\; dx\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80 x \bigg]_{80}^{90}\\ &= \frac{1}{800}\bigg[\frac{90^2}{2} - (80\times 90) - \frac{80^2}{2} +(80\times 80)\bigg]\\ &=0.0625 \end{aligned}

b. The mean resistance is

 \begin{aligned} E(X) &= \int_{80}^{120} xf(x)\; dx\\ &= \int_{80}^{120} x\frac{x-80}{800}\; dx\\ &= \frac{1}{800}\int_{80}^{120} (x^2-80x)\; dx\\ &= \frac{1}{800}\bigg[\frac{x^3}{3} - \frac{80x^2}{2} \bigg]_{80}^{120}\\ &= \frac{1}{800}\bigg[\frac{120^3}{3} - (80\times \frac{120^2}{2}) - \frac{80^3}{3} +(80\times\frac{80^2}{2})\bigg]\\ &=106.667 \end{aligned}

c. For standard deviation, let us calculate $E(X^2)$.

 \begin{aligned} E(X^2) &= \int_{80}^{120} x^2f(x)\; dx\\ &= \int_{80}^{120} x^2\frac{x-80}{800}\; dx\\ &= \frac{1}{800}\int_{80}^{120} (x^3-80x^2)\; dx\\ &= \frac{1}{800}\bigg[\frac{x^4}{4} - \frac{80x^3}{3} \bigg]_{80}^{120}\\ &= \frac{1}{800}\bigg[\frac{120^4}{4} - (80\times \frac{120^3}{3}) - \frac{80^4}{4} +(80\times\frac{80^3}{3})\bigg]\\ &=11466.67 \end{aligned}

Thus, $V(X) = E(X^2) - [E(X)]^2 = 11466.67 - (106.667)^2 =88.821$

The standard deviation of the resistances is $sd(X) =\sqrt{V(X)} =\sqrt{88.821} = 9.4245$.

d. The cumulative distribution function of the resistances is

 \begin{aligned} F(x) &= P(X\leq x)\\ &=\int_{80}^x f(x)\; dx\\ &= \frac{1}{800}\int_{80}^x (x-80)\; dx\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x \bigg]_{80}^x\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x -\frac{80^2}{2} + 80\times 80\bigg]\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x +3200\bigg] \end{aligned}