The amount of time, in hours, that a machine functions before breakdown is a continuous random variable with pdf

` $$ \begin{aligned} f(t) &= \frac{1}{120} e^{-t/120}, t\geq 0 \end{aligned} $$ `

What is the probability that this machine will function between 98 and 145 h before breakdown? What is the probability that it will function less than 160 h?

#### Solution

The distribution function of $t$ is

` $$ \begin{aligned} F(t) &= 1- e^{-t/120}. \end{aligned} $$ `

The probability that this machine will function between 98 and 145 h before breakdown is

` $$ \begin{aligned} P(98 < t < 145) &= P(t\leq 145) - P(t \leq 98)\\ & = F(145) - F(98)\\ & = \big(1- e^{-145/120}\big) - \big(1- e^{-98/120}\big)\\ &= e^{-98/120} -e^{-145/120}\\ & = 0.4419022 - 0.2986947\\ &= 0.1432075 \end{aligned} $$ `

The probability that this machine will function less than 160 h is

` $$ \begin{aligned} P(t < 160) &= P(t\leq 160)\\ & = F(160)\\ & = \big(1- e^{-160/120}\big) \\ & = 1- 0.2635971 \\ &= 0.7364029 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators