Question

Persons Per Housing Unit. The random variable Y is the number of person living in a randomly selected occupied housing unit. Its probability distribution is as follows.

Part (a) Find and interpret the mean of the random variable.

Part (b) Obtain the standard deviation of the random variable by using one of the formulas given in definition 5.10

Part (c) Construct a probability histogram for the random variable, locate the mean: and show one, two, and three standard deviation intervals.

Short answer

Part (a) 2.538

Part (b) 1.4

Part (c)

Step-by-step solution

Part (a) Step 1. Given information. 

Consider the following table of data. The random variable Y is the number of persons living in randomly selected household units.

Part (a) Step 2. The random variable's average.

$$\begin{gathered} \mu =\sum _{}y·P(Y=y) \\ \mu =1\left(0.365\right)+\left(2\times 0.0327\right)+\left(3\times 0.161\right)+\left(4\times 0.147\right)+\left(5\times 0.065\right)+6\left(0.022\right)+7\left(0.013\right) \\ \mu =0.265+0.654+0.483+0.588+0.326+0.132+0.091 \\ \mu =2.538 \end{gathered}$$

The average number of people living in a particular occupied dwelling unit is = 2.538.

Part (b) Step 1.The random variable's standard deviation:

Standard deviation = $\sqrt{\sum _{}{y}^{2}P(Y=y)-{\mu }^2}$

Calculation.

$$\begin{gathered} \sigma =\sqrt{\sum _{}{x}^{2}P(X=x)-{\mu }^2} \\ \sigma =\sqrt{1\left(0.365\right)+\left(4\times 0.327\right)+\left(9\times 0.161\right)+\left(16\times 0.147\right)+\left(25\times 0.065\right)-36.(0.022)+49(0.013)-{(2.53)}^2} \\ \sigma =\sqrt{1.987} \\ \sigma =1.4 \end{gathered}$$

As a result, standard deviation = 1.4

Part (c) Step 1.  The mean, one, two, and three standard deviation intervals on the probability histogram graph of the random variable.

We have $\mu$and $\sigma$from sections a and b.

localid="1651756716707" $$\begin{gathered} \left(\mu -\sigma ,\mu +\sigma \right)=\left(2.538-1.4,2.538+1.4\right) \\ =\left(1.138,3.938\right) \\ \left(\mu -2\sigma ,\mu +2\sigma \right)=\left(2.538-2(1.4),2.538+2(1.4)\right) \\ =(-0.2,5.3) \\ \left(\mu -3\sigma ,\mu +3\sigma \right)=\left(2.538-3(1.4),2.538+3(1.4)\right) \\ (-1.6,6.7) \end{gathered}$$

Draw the probability histogram graph: