Chapter 6.1, Problem 44E

Do you carpool? Let X represent the number of occupants in a randomly chosen car on a certain stretch of highway during morning commute hours. A survey of cars showed that the probability distribution of X is as follows.

1. Find P(2).
2. Find P(No more than 1).
3. Find the probability that at least two circuits are defective.
4. Find the probability that no more than two circuits are defective.
5. Compute the mean $\mu X.$
6. Compute the standard deviation $\sigma X.$
7. A circuit will function if it has no defects or only one defect. What is the probability that a circuit will function?

To determine

To find:The probability of the random variable $x=2$ .

Answer to Problem 44E

The probability of $x=2$ is found to be,

  $P\left(2\right)=0.15$

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following probability distribution.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

By the notation $P\left(2\right)$ the probability to obtain $2$ occupants is denoted. When $x=2$ the corresponding probability is given in the distribution as $0.15$ .

Therefore,

  $P\left(x=2\right)=0.15$

Conclusion:

The probability to obtain two occupants in a randomly selected car is found to be $0.15$ .

To determine

To find:The probability to the random variable to be more than $3$ .

Answer to Problem 44E

The probability $P\left(\text{More than }3\right)$ is found to be $0.09$ .

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following probability distribution.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

The random variable $x$ is more than $3$ says that $x>3$ . Hence, the values that the variable can approach are $4$ and $5$ as given in the distribution.

Hence, the probability $P\left(X>3\right)$ can be obtained by $P\left(4\text{ or }5\right)$ . Obtaining $4$ occupants and $5$ occupants are two mutually exclusive events. Therefore, by the addition rule of mutually exclusive events,

  $\begin{array}{c}P\left(4\text{ or }5\right)=P\left(4\right)+P\left(5\right)\\ =0.03+0.01\\ P\left(4\text{ or }5\right)=0.04\end{array}$

Conclusion:

The probability to obtain more than $3$ occupants is found to be,

  $P\left(x>3\right)=0.04$

To determine

To find:The probability to obtain only one occupant in a randomly selected car.

Answer to Problem 44E

The probability that a car has only one occupant is found to be,

  $P\left(1\right)=0.70$

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following probability distribution.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

The distribution shows the values of the random variable and the corresponding probabilities.

When a car has only one occupant, the random variable is said to be $x=1$ . Hence, the probability to be $x=1$ is given as,

  $P\left(x=1\right)=0.70$ .

Conclusion:

The probability to have one occupant in a car is found to be $0.70$ .

To determine

To find:The probability to obtain few occupants than four in a randomly selected car.

Answer to Problem 44E

  $0.95$ is the probability to have fewer occupants than four.

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following table.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

As the condition to have fewer occupants than four, the random variable $x$ approaches to the values of $1,2$ and $3$ .

Hence, the probability can be written as,

  $P\left(x<4\right)=P\left(1\text{ or }2\text{ or }3\right)$

Since, all these three events $x=1,x=2$ and $x=3$ are mutually exclusive events,

  $\begin{array}{c}P\left(1\text{ or }2\text{ or }3\right)=P\left(1\right)+P\left(2\right)+P\left(3\right)\\ =0.70+0.15+0.10\\ =0.95\end{array}$

Conclusion:

The probability of $P\left(x<4\right)$ is found to be $0.95$ .

To determine

To find:The mean occupants in the cars.

Answer to Problem 44E

The mean is found to be,

  ${\mu }_X=1.52$

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following table.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

The mean of a random variable, or equivalently the expected value is given by the sum of the product of the values and the corresponding probabilities.

  ${\mu }_X=E\left(X\right)=x_1{P}_1+x_2{P}_2+x_3{P}_3+\mathrm{...}$

Here, for each value of $x$ should be multiplied by the probabilities and added.

  $\begin{array}{c}{\mu }_X=\left(1\times 0.70\right)+\left(2\times 0.15\right)+\left(3\times 0.10\right)+\left(4\times 0.03\right)+\left(5\times 0.02\right)\\ =0.70+0.30+0.30+0.12+0.10\\ {\mu }_X=1.52\end{array}$

Conclusion:

The mean number of occupants in the cars is found to be $1.52$ .

To determine

To find:The standard deviation of $X$ .

Answer to Problem 44E

The standard deviation is found to be,

  ${\sigma }_X=0.9325$

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following table.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

Calculation:

The variance of a random variable $\left({\sigma }_X{}^2\right)$ is given by the formula,

  ${\sigma }_X{}^2={{\displaystyle \sum \left[\left(x-{\mu }_X\right)\cdot P\left(x\right)\right]}}^2$$$

By constructing a table we can do the calculations clearly.

  $\begin{array}{ccccc}x& x-{\mu }_X& {\left(x-{\mu }_X\right)}^2& P\left(x\right)& {\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)\\ 1& -0.52& 0.2704& 0.70& 0.1893\\ 2& 0.48& 0.2304& 0.15& 0.0346\\ 3& 1.48& 2.1904& 0.10& 0.2190\\ 4& 2.48& 6.1504& 0.03& 0.1845\\ 5& 3.48& 12.1104& 0.02& 0.2422\end{array}$

The sum of right-most column gives the variation of $X$ .

  ${\sigma }_X{}^2=0.8696$

The standard deviation $\left({\sigma }_X\right)$ is the square root of variance. Hence,

  $\begin{array}{c}{\sigma }_X=\sqrt{{\sigma }_X{}^2}\\ =\sqrt{0.8696}\\ {\sigma }_X=0.9325\end{array}$

Conclusion:

The standard deviation is found to be $0.9325$ .

To determine

To check:Whether the goal of saving energy has been achieved.

Explanation of Solution

The random variable $X$ represents the number of occupants of randomly selected car on a stretch of highway. The probability for $X=\left\{1,2,3,4,5\right\}$ is given in the following table.

  $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ P\left(x\right)& 0.70& 0.15& 0.10& 0.03& 0.02\end{array}$

The goal to save energy is set as the mean number of occupants should be less than $2$ . As calculated in the part (e), the mean number of occupants is said to be $1.52$ .

Since, $1.52<2$ , the condition of the goal has been satisfied.

Therefore, we can conclude that the goal to save energy has been met.