MATH SOLVE

2 months ago

Q:
# The waiting time, in hours, between successive speeders spotted by a radar unit is a continuous random variable with cumulative distribution function f(x) = 0, x< 0, 1 − e −8x, x≥ 0. find the probability of waiting less than 12 minutes between successive speeders (a) using the cumulative distribution function of x; (b) using the probability density function of x.

Accepted Solution

A:

Given the CDF

[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\1-e^{-8x}&\text{for }x\ge0\end{cases}[/tex]

we have PDF

[tex]\dfrac{\mathrm dF_X(x)}{\mathrm dx}=f_X(x)=\begin{cases}0&\text{for }x<0\\8e^{-8x}&\text{for }x\ge0\end{cases}[/tex]

Note that [tex]X[/tex] is wait-time given in hours, so we need to convert from minutes to hours:

[tex]12\text{ min}\times\dfrac{1\text{ hr}}{60\text{ min}}=\dfrac15\text{ hr}[/tex]

so we're looking for [tex]\mathbb P\left(X<\dfrac15\right)[/tex].

The CDF gives us this value right away, since [tex]F_X(x)=\mathbb P(X<x)=\mathbb P(X\le x)[/tex] for any continuous random variable [tex]X[/tex] with distribution function [tex]F_X(x)[/tex]:

[tex]\mathbb P\left(X<\dfrac15\right)=F_X\left(\dfrac15\right)=1-e^{-8/5}\approx0.7981[/tex]

To use the PDF, we need to integrate:

[tex]\mathbb P\left(X<\dfrac15\right)=\displaystyle\int_{-\infty}^{1/5}f_X(x)\,\mathrm dx=\int_0^{1/5}8e^{-8x}\,\mathrm dx\approx0.7981[/tex]

[tex]F_X(x)=\begin{cases}0&\text{for }x<0\\1-e^{-8x}&\text{for }x\ge0\end{cases}[/tex]

we have PDF

[tex]\dfrac{\mathrm dF_X(x)}{\mathrm dx}=f_X(x)=\begin{cases}0&\text{for }x<0\\8e^{-8x}&\text{for }x\ge0\end{cases}[/tex]

Note that [tex]X[/tex] is wait-time given in hours, so we need to convert from minutes to hours:

[tex]12\text{ min}\times\dfrac{1\text{ hr}}{60\text{ min}}=\dfrac15\text{ hr}[/tex]

so we're looking for [tex]\mathbb P\left(X<\dfrac15\right)[/tex].

The CDF gives us this value right away, since [tex]F_X(x)=\mathbb P(X<x)=\mathbb P(X\le x)[/tex] for any continuous random variable [tex]X[/tex] with distribution function [tex]F_X(x)[/tex]:

[tex]\mathbb P\left(X<\dfrac15\right)=F_X\left(\dfrac15\right)=1-e^{-8/5}\approx0.7981[/tex]

To use the PDF, we need to integrate:

[tex]\mathbb P\left(X<\dfrac15\right)=\displaystyle\int_{-\infty}^{1/5}f_X(x)\,\mathrm dx=\int_0^{1/5}8e^{-8x}\,\mathrm dx\approx0.7981[/tex]