Chapter 6.1, Problem 42E

Fifteen items or less: The number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution.

Find P(2).

Find P(No more than 1).

Find the probability that no one is in line.

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

Compute the mean $\mu X.$

Compute the standard $\sigma X.$

If each customer takes 3 minutes to check out: what is the probability that it will take more than 6 minutes for all the customers currently in line to check out?

To determine

To find: $P(2)$

Explanation of Solution

Given information:number of customers atcheckout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$ .

  

the value of $P(2)$ can be read as shown.

We locate $x=2$ along the $x$ axis and move vertically upward until the height for $P(x)$ is reached. From this point on the line, we move horizontally to the left until the $y$ axis is reached. Reading the $P(2)$ , we find that the $P(2)=0.30$ .

Therefore,

  $P(2)=0.30$

To determine

To find: $P($ No more than $1)$

Explanation of Solution

Given information:number of customers atcheckout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

the values of $P(x\le 1)$ can be read as shown.

We locate $x=0$ and $x=1$ along the $x$ axis and move vertically upward until the height for $P(x)$ is reached. From this point on the line, we move horizontally to the left until the $y$ axis is reached. Reading the $P(0)$ and $P(1)$ , we find that the $P(0)=0.10$ and $P(1)=0.25$ .

Therefore,

  $\begin{array}{l}P(x\le 1)=P(0)+P(1)\\ P(x\le 1)=0.10+0.25\\ P(x\le 1)=0.35\end{array}$

The $P(x\le 1)$ includes both $P(0)$ and $P(1)$ .Because it is for $x$ values not more than $1$ .

To determine

To find: the probability no one is the line.

Explanation of Solution

Given information:number of customers atcheckout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

the values of $P(0)$ can be read as shown.

We locate $x=0$ along the $x$ axis and move vertically upward until the height for $P(x)$ is reached. From this point on the line, we move horizontally to the left until the $y$ axis is reached. Reading the $P(0)$ , we find that the $P(0)=0.10$ .

  $P(0)=0.10$

Therefore,the probability no one is the line is given by $P(0)$ .

To determine

To find: the probability at least three people are in the line

Explanation of Solution

Given information:number of customers atcheckout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

the values of $P(x\ge 3)$ can be read as shown.

We locate $x=3$ , $x=4$ , $x=5$ along the $x$ axis and move vertically upward until the height for $P(x)$ is reached. From this point on the line, we move horizontally to the left until the $y$ axis is reached. Reading the $P(3),P(4),P(5)$ , we find that the $P(3)=0.20$ , $P(4)=0.10$ and $P(5)=0.05$ .

Therefore,the probability that at least three people are in the line is the addition of $P(3)=0.20$ , $P(4)=0.10$ and $P(5)=0.05$ .

  $\begin{array}{l}P(x\ge 3)=P(3)+P(4)+P(5)\\ P(x\ge 3)=0.20+0.10+0.05\\ P(x\ge 3)=0.35\end{array}$

To determine

To calculate: mean ${\mu }_x$

Explanation of Solution

Given information:number of customers atcheckout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

mean ${\mu }_x$ can be calculated using the formula ${\mu }_x={\displaystyle \sum x.P(x)}$ as shown.

Therefore,

  $\begin{array}{l}{\mu }_x={\displaystyle \sum x.P(x)}\\ {\mu }_x=(0)(0.10)+(1)(0.25)+(2)(0.30)+(3)(0.20)+(4)(0.10)+(5)(0.05)\\ {\mu }_x=0+0.25+0.60+0.60+0.40+0.25\\ {\mu }_x=2.1\end{array}$

To determine

To calculate: standard deviation ${\sigma }_x$

Explanation of Solution

Given information: number of customers at checkout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

standard deviation ${\sigma }_x$ can be calculated using the formula ${\sigma }_x^2={\displaystyle \sum [x^2}P(x)]-{\mu }_x^2$ as shown.

Therefore,

  $\begin{array}{l}{\sigma }_x^2={\displaystyle \sum [x^2}P(x)]-{\mu }_x^2\\ {\sigma }_x^2=(0)(0.10)+(1)(0.25)+(4)(0.35)+(9)(0.20)+(16)(0.10)+(25)(0.05)-{(}^{2.1}\\ {\sigma }_x^2=0+0.25+1.40+1.80+1.60+1.25-4.41\\ {\sigma }_x^2=6.30-4.41\\ {\sigma }_x^2=1.89\\ {\sigma }_x=1.37477270849\\ {\sigma }_x=1.37\end{array}$

To determine

To find: the probability that takes more than six minutes (per customer it is 3 minutes) for all the customers in line to check out

Explanation of Solution

Given information: number of customers at checkout counter is a random variable with the following probability distribution.

  $\begin{array}{cccccccccccccccccc}x& 0& 1& 2& 3& 4& 5& & & & & & & & & & & \\ P(x)& 0.10& 0.25& 0.30& 0.20& 0.10& 0.05& & & & & & & & & & & \end{array}$

Graph:the line graph shows $P(x)$ vs. $x$

  

the values of $P(x\ge 2)$ can be read as shown.

We locate $x=2,3,4,5$ along the $x$ axis and move vertically upward until the height for $P(x)$ is reached. From this point on the line, we move horizontally to the left until the $y$ axis is reached. Reading the $P(2),P(3),P(4),P(5)$ , we find that the $P(2)=0.30,P(3)=0.20,P(4)=0.10,P(5)=0.05$

Therefore,the probability that takes more than six minutes (per customer it is 3 minutes) for all the customers in line to check out is,

  $\begin{array}{l}P(x\ge 2)=P(2)+P(3)+P(4)+P(5)\\ P(x\ge 2)=0.30+0.20+0.10+0.05\\ P(x\ge 2)=0.65\end{array}$