Chapter 5.1, Problem 42E

a.

To determine

Find the value of $P\left(2\right)$.

Answer to Problem 42E

The value of $P\left(2\right)$ is 0.30.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the number of customers in a line at a supermarket express checkout counter.

From the probability distribution table, the probability at the point $x=2$ is $P\left(2\right)=0.30$.

Thus, the value of $P\left(2\right)$ is 0.30.

b.

To determine

Find the probability value, $P\left(\text{No more than 1}\right)$.

Answer to Problem 42E

The probability value, $P\left(\text{No more than 1}\right)$ is 0.35.

Explanation of Solution

Calculation:

The required probability is the sum of the probabilities at $x=0\text{ and }x=1$.

$\begin{array}{c}P\left(\text{No more than 1}\right)=P\left(x\le 1\right)\\ =P\left(x=0\right)+P\left(x=1\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{No more than 1}\right)=P\left(x=0\right)+P\left(x=1\right)\\ =0.10+0.25\\ =0.35\end{array}$

Thus, the probability value, $P\left(\text{No more than 1}\right)$ is 0.35.

c.

To determine

Find the probability that no one is in line.

Answer to Problem 42E

The probability that no one is in line is 0.10.

Explanation of Solution

Calculation:

The probability that no one is in line is the probability at the point $x=0$. From the probability distribution table, the value of $P\left(0\right)$ is 0.10.

Thus, the probability that no one is in line is 0.10.

d.

To determine

Find the probability that at least three people are in line.

Answer to Problem 42E

The probability that at least three people are in line is 0.35.

Explanation of Solution

Calculation:

The probability that at least three people are in line is the sum of the probabilities at $x=3,x=4\text{ and }x=5$.

$\begin{array}{c}P\left(\text{At least 3}\right)=P\left(x\ge 3\right)\\ =P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\end{array}$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{At least 3}\right)=P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\\ =0.20+0.10+0.05\\ =0.35\end{array}$

Thus, the probability that at least three people are in line is 0.35.

e.

To determine

Find the mean.

Answer to Problem 42E

The mean value is 2.10.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

$\begin{array}{c}E\left(X\right)={\mu }_X\\ ={\displaystyle \sum x\cdot P\left(x\right)}\end{array}$

The mean of the random variable is obtained as given below:

x	P(x)	$x\cdot P\left(x\right)$
0	0.10	0.0
1	0.25	0.3
2	0.30	0.6
3	0.20	0.6
4	0.10	0.4
5	0.05	0.3
Total	1.00	2.1

Thus, the mean value is 2.10.

f.

To determine

Find the standard deviation.

Answer to Problem 42E

The standard deviation is 1.30.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

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

Where ${\mu }_X={\displaystyle \sum \left[x\cdot P\left(x\right)\right]}$ represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

x	P(x)	$\left(x-{\mu }_X\right)$	${\left(x-{\mu }_X\right)}^2$	${\left(x-{\mu }_X\right)}^2\cdot P\left(x\right)$
0	0.10	–2.1	4.41	0.44
1	0.25	–1.1	1.21	0.30
2	0.30	–0.1	0.01	0.00
3	0.20	0.9	0.81	0.16
4	0.10	1.9	3.61	0.36
5	0.05	2.9	8.41	0.42
Total		2.4	18.46	1.69

Therefore,

${\sigma }^2=1.69$

Thus, the variance is 1.69.

The standard deviation is,

$\begin{array}{c}\sigma =\sqrt{1.69}\\ =1.30\end{array}$

That is, the standard deviation is 1.30.

g.

To determine

Find the probability that it will take more than 6 minutes for all the customers currently in line to checkout.

Answer to Problem 42E

The probability that it will take more than 6 minutes for all the customers currently in line to checkout is 0.35.

Explanation of Solution

Calculation:

Each customer will take three minutes to check out. If the number of customers is 1 then it will take 3 minutes, if the number of customers is 2 it will take 6 minutes, if the number of customers is 3 or more then it will take more than 6 minutes.

Thus, the probability that that it will take more than 6 minutes for all the customers currently in line to checkout is the sum of the probabilities at $x=3,x=4\text{ and }x=5$.

$P\left(\text{More than 6 minutes to checkout}\right)=P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)$

Substituting the values from the probability distribution table,

$\begin{array}{c}P\left(\text{More than 6 minutes to checkout}\right)=P\left(x=3\right)+P\left(x=4\right)+P\left(x=5\right)\\ =0.20+0.10+0.05\\ =0.35\end{array}$

Thus, the probability that it will take more than 6 minutes for all the customers currently in line to checkout is 0.35.