Chapter 5.4, Problem 121AYU

(a)

To determine

The number of minutes needed for the probability to reach $50\%$ , where between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes).The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$ .

Answer to Problem 121AYU

Solution:

Approximately, within $7\text{minutes}$ of 12:00 PM the probability reaches to $50\%$

Explanation of Solution

Given information:

Between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes).

The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$

Here, probability is $50\%$

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

It gives $50\%=1-e^{-0.1t}$

By simplifying,

  $\Rightarrow \frac{1}{2}=1-e^{-0.1t}$

Subtract $1$ from both sides,

  $\Rightarrow \frac{1}{2}-1=1-e^{-0.1t}-1$

  $\Rightarrow -\frac{1}{2}=-e^{-0.1t}$

  $\Rightarrow \frac{1}{2}=e^{-0.1t}$

Taking natural log on both sides,

  $\Rightarrow \ln \left(\frac{1}{2}\right)=\ln \left(e^{-0.1t}\right)$

  $\Rightarrow \ln \left(\frac{1}{2}\right)=-0.1t$

  $\Rightarrow t=6.93147\approx 7$

Therefore, approximately $7$ minutes needed for the probability to reach $50\%$

(b)

To determine

The number of minutes needed for the probability to reach $80\%$ , where between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes).The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$ .

Answer to Problem 121AYU

Solution:

Approximately, within $16$minutes of 12:00 PM probability reaches to $80\%$

Explanation of Solution

Given information:

Between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes)

The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$

Here, probability is $80\%$

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

It gives $80\%=1-e^{-0.1t}$

By simplifying,

  $\frac{8}{10}=1-e^{-0.1t}$

Subtract $1$ from both sides,

  $\Rightarrow \frac{8}{10}-1=1-e^{-0.1t}-1$

  $\Rightarrow -\frac{2}{10}=-e^{-0.1t}$

  $\Rightarrow \frac{1}{5}=e^{-0.1t}$

Taking natural log on both sides,

  $\Rightarrow \ln \left(\frac{1}{5}\right)=\ln \left(e^{-0.1t}\right)$

  $\Rightarrow \ln \left(\frac{1}{5}\right)=-0.1t$

  $\Rightarrow t=16.09437\approx 16$

Therefore, approximately $16$ minutes needed for the probability to reach $80\%$

(c)

To determine

Whether the probability equals $100\%$ or not, where between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes).The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$

Answer to Problem 121AYU

Solution:

The probability equals $100\%$ is not possible.

Explanation of Solution

Given information:

Between $12:00\text{PM}$ and $1:00\text{PM}$ , cars arrive at Citibank’s drive thru at the rate of $6$ cars per hour( $0.1$ car per minutes).

The probability that a car will arrive within $t$ minutes of $12:00\text{PM}$ is $F\left(t\right)=1-e^{-0.1t}$

If the probability is $100\%$

Thus, $F\left(t\right)=100\%$

It gives, $100\%=1-e^{-0.1t}$

By simplifying,

  $1=1-e^{-0.1t}$

Subtract $1$ from both sides,

  $\Rightarrow 1-1=1-e^{-0.1t}-1$

  $\Rightarrow 0=-e^{-0.1t}$

  $\Rightarrow e^{-0.1t}=0$ , it has no solution.

Since the exponential is never zero

Thus, there does not exist $t$ for which the probability is $100\%$

Therefore, the probability equals $100\%$ is not possible.