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
$$ \begin{aligned} F(x) &= P(X\leq x)\\ &=\int_{80}^x f(x)\; dx\\ &= \frac{1}{800}\int_{80}^x (x-80)\; dx\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x \bigg]_{80}^x\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x -\frac{80^2}{2} + 80\times 80\bigg]\\ &= \frac{1}{800}\bigg[\frac{x^2}{2} - 80x +3200\bigg] \end{aligned} $$
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
