Chapter 5.4, Problem 122AYU

(a)

To determine

The number of minutes needed for the probability to reach $50\%$, where between $5:00\text{PM}$ and $6:00\text{PM}$ cars arrive at Jiffy Lube at the rate of $9$ cars per hour ( $0.15$ car per minutes). The probability that a car will arrive within $t$ minutes of $5:00\text{PM}$ is $F\left(t\right)=1-e^{-0.15t}$

Answer to Problem 122AYU

Solution:

Approximately, within $5$ minutes of 5:00 PM the probability reaches to $50\%$

Explanation of Solution

Given information:

Between $5:00\text{PM}$ and $6:00\text{PM}$, cars arrive at Jiffy Lube at the rate of $9$ cars per hour ( $0.15$ car per minutes).

The probability that a car will arrive within $t$ minutes of $5:00\text{PM}$ is $F\left(t\right)=1-e^{-0.15t}$

Explanation:

Here, the probability is $50\%$

Thus, $F\left(t\right)=50\%$

$\Rightarrow 50\%=1-e^{-0.15t}$

By simplifying,

$\Rightarrow \frac{1}{2}=1-e^{-0.15t}$

Subtract $1$ from both sides,

$\Rightarrow \frac{1}{2}-1=1-e^{-0.15t}-1$

$\Rightarrow -\frac{1}{2}=-e^{-0.15t}$

$\Rightarrow \frac{1}{2}=e^{-0.15t}$

Taking natural log on both sides,

$\Rightarrow \ln \left(\frac{1}{2}\right)=\ln \left(e^{-0.15t}\right)$

$\Rightarrow \ln \left(\frac{1}{2}\right)=-0.15t$

$\Rightarrow t=4.62098\approx 5$

Therefore, approximately $5$ minutes needed for the probability to reach $50\%$

(b)

To determine

The number of minutes needed for the probability to reach $80\%$, where between $5:00\text{PM}$ and $6:00\text{PM}$, cars arrive at Jiffy Lube at the rate of $9$ cars per hour ( $0.15$ car per minutes). The probability that a car will arrive within $t$ minutes of $5:00\text{PM}$ is $F\left(t\right)=1-e^{-0.15t}$

Answer to Problem 122AYU

Solution:

Approximately, within $11$ minutes of 5:00 PM the probability reaches to $80\%$

Explanation of Solution

Given information:

Between $5:00\text{PM}$ and $6:00\text{PM}$, cars arrive at Jiffy Lube at the rate of $9$ cars per hour ( $0.15$ car per minutes).

The probability that a car will arrive within $t$ minutes of $5:00\text{PM}$ is $F\left(t\right)=1-e^{-0.15t}$

Explanation:

Here, probability is $80\%$

Thus, $F\left(t\right)=80\%$

$\Rightarrow 80\%=1-e^{-0.15t}$

By simplifying,

$\Rightarrow \frac{8}{10}=1-e^{-0.15t}$

Subtract $1$ from both sides,

$\Rightarrow \frac{8}{10}-1=1-e^{-0.15t}-1$

$\Rightarrow -\frac{2}{10}=-e^{-0.15t}$

$\Rightarrow \frac{1}{5}=e^{-0.15t}$

Taking natural log on both sides,

$\Rightarrow \ln \left(\frac{1}{5}\right)=\ln \left(e^{-0.15t}\right)$

$\Rightarrow \ln \left(\frac{1}{5}\right)=-0.15t$

$\Rightarrow t=10.72958\approx 11$

Therefore, approximately $11$ minutes needed for the probability to reach $80\%$