Queuing Simulation. Benny's Barber Shop in Cleveland has five chairs for waiting customers. The number of customers waiting is a random variable $Y$with the following probability distribution.

mass placed at point $x_k$on the seesaw. The center of gravity of these masses is defined to be the point $c$on the horizontal axis at which a fulcrum could be placed to balance the seesaw.

Relative to the center of gravity, the torque acting on the seesaw by the mass $p_k$is proportional to the product of that mass with the signed distance of the point $x_k$from c, that is, to $\left(x_k-c\right)·p_k$. Show that the center of gravity equals the mean of the random variable$X$. (Hint: To balance, the total torque acting on the seesaw must be 0 .)

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

${\mu }_{\gamma }=\sum yP(Y=y)$

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

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

$=1.519$

$\approx 2$

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:

${\mu }_{y^{*}}=\sum y^{\text{'}}P\left(Y^{\text{'}}=y^{\text{'}}\right)$

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

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

$=2.516$

$\approx 3$