Question

Red light! Pedro drives the same route to work on Monday through Friday. His route includes one traffic light. According to the local traffic department, there is a $55\%$chance

that the light will be red on a randomly selected work day. Suppose we choose 10 of Pedro's work days at random and let Y= the number of times that the light is red. Make a graph of the probability distribution of Y . Describe its shape.

Short answer

The data seems to be symmetrical

Step-by-step solution

Step 1:Given Information

Given information.

The probability of the light turning red on a random workday is $55$percent $=0.55$.

The total number of working days is ten.

Simplification

Let $Y$represent the number of times the light turns red.

Because the variable$Y$satisfies the following properties, it follows a binomial distribution.

1. The number of independent trials

2. Each trial has an equal chance of success or failure.

3. The outcome of each trial is either success or failure.

4. The number of trials is predetermined.

The probability function of $Y$is $P\left(y\right)=\left(\begin{array}{c}10\\ y\end{array}\right)(0.55{)}^y(1-0.55{)}^{10-y}$

Insert the values of $y=0,1,2,\dots ,12$in the probability function to get the individual probabilities.

The probability of no red light on $x=0$is,

$\begin{array}{rcl}P\left(0\right)& =& \left(\begin{array}{l}10\\ 0\end{array}\right)(0.55{)}^0(0.45{)}^{10-0}\\ & =& 0.00034\end{array}$

The possibility of a one-day red light on$x=1$is,

$$\begin{gathered} P\left(1\right)=\left(\begin{array}{l}10\\ 1\end{array}\right)(0.55{)}^1(0.45{)}^{10-1} \\ =120(0.55)(0.45{)}^8 \\ =0.00416 \end{gathered}$$

The probability that two days red light on $x=2$is,

$\begin{array}{rcl}P\left(2\right)& =& \left(\begin{array}{c}10\\ 2\end{array}\right)(0.55{)}^2(0.45{)}^{10-2}\\ & =& \frac{12\times 11}{2\times 1}(0.55{)}^2(0.45{)}^8\\ & =& 0.0229\end{array}$

Continue in the same manner to obtain the probabilities.

Further calculation

The probability distribution of the variable Y is shown below.

$y$	$p\left(y\right)$
0	localid="1657633858647" $0.0034$
1	localid="1657633861317" $0.00416$
localid="1657633855711" $2$	localid="1657633864046" $0.02289$
localid="1657633881452" $3$	$0.0746$
localid="1657633884843" $4$	$0.15957$
localid="1657633888586" $5$	localid="1657633867369" $0.23403$
localid="1657633893821" $6$	localid="1657633870761" $0.23837$
localid="1657633897012" $7$	localid="1657633873538" $0.16648$
localid="1657633900334" $8$	localid="1657633876780" $0.0763$
localid="1657633903531" $9$	localid="1657633918004" $0.02072$
width="19" style="max-width: none; vertical-align: -4px;" $10$	$0.00253$

The probability distribution graph is shown below.

The distribution appears to be roughly symmetric. The greatest likelihood occurs when $x=6$.