91Ó°ÊÓ

By definitions, the mean of the random variable $Y$is given as follows:

${\mathrm{μ}}_{\mathrm{γ}}=∑yP(Y=y)$

$=\left[\right(0×0.424)+(1×0.161)+…+(5×0.077\left)\right]$

$=(0.000+0.161+0.268+0.333+0.372+0.385)$

$=1.519$

$≈2$

Interpretation: Therefore, on an average, number of customers will wait at a given time are 2

In a large number of independent observations, on an average, the number of customers will be in waiting is $2$because the size of the sample doesn't change the probability of the outcome.

Using MINITAB, we simulate 500 observations of the number of customers waiting and thus we obtain the frequency distribution of the above simulated values as follows:

From part (c), we obtain the relative frequencies for the simulated frequency distribution as follows:

As the variable number of customers waiting is a variable of finite population, its probabilities are identical to its relative frequencies.

By definitions, the mean of the observations obtained in part (c) is given as follows:

${\mathrm{μ}}_{y^{*}}=∑y^{\text{'}}P\left(Y^{\text{'}}=y^{\text{'}}\right)$

$=\left[\right(0×0.116)+(1×0.158)+…+(5×0.164\left)\right]$

$=(0.000+0.158+0.344+0.498+0.696+0.820)$

$=2.516$

$≈3$