91Ó°ÊÓ

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

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

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

Standard deviation = $\sqrt{\underset{}{∑}{y}^{2}P(Y=y)-{\mathrm{μ}}^2}$

Calculation.

$$\begin{gathered} \mathrm{σ}=\sqrt{\underset{}{∑}{x}^{2}P(X=x)-{\mathrm{μ}}^2} \\ \mathrm{σ}=\sqrt{1\left(0.365\right)+\left(4×0.327\right)+\left(9×0.161\right)+\left(16×0.147\right)+\left(25×0.065\right)-36.(0.022)+49(0.013)-{(2.53)}^2} \\ \mathrm{σ}=\sqrt{1.987} \\ \mathrm{σ}=1.4 \end{gathered}$$

As a result, standard deviation = 1.4

We have $\mathrm{μ}$and $\mathrm{σ}$from sections a and b.

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

Draw the probability histogram graph: