Analysis of supplier lead time. Lead timeis the time betweena retailer placing an order and having the productavailable to satisfy customer demand. It includes time for placing the order, receiving the shipment from the supplier, inspecting the units received, and placing them in inventory. Interested in average lead time,, for a particular supplier of men’s apparel, the purchasing department of a national department store chain randomly sampled 50 of the supplier’s lead times and found= 44 days.

1. Describe the shape of the sampling distribution of$\overline{x}$.
2. If $\mu$and $\sigma$are really 40 and 12, respectively, what is the probability that a second random sample of size 50 would yield$\overline{x}$ greater than or equal to 44?
3. Using the values for$\mu$ and $\sigma$in part b, what is the probability that a sample of size 50 would yield a sample mean within the interval $\mathbf{\mu }\mathbf{\pm }\frac{\mathbf{2}\mathbf{\sigma }}{\sqrt{\mathbf{n}}}$?

There is a lead time between a retailer placing an order. The lead time includes the times for placing the order, receiving the shipment from the supplier, inspecting the received unit, and placing it into the inventory.

For a particular supplier who supplies men’s apparel, the purchasing department of a national department store chain randomly sampled 50 of the supplier’s lead times and found the sample mean $\overline{x}=44$.

a.

Let’s consider that the random variable X be the lead time of the suppliers. So, the sample mean is defined as the average of the lead time of the suppliers that is $\overline{x}$.

Given, the mean and the standard deviation of the lead times are 44 and 12 respectively.

The sampling distribution of$\overline{x}$ follows a normal distribution with a mean 44 and the standard deviation of

$\begin{array}{rcl}\frac{\sigma }{\sqrt{n}}& =& \frac{12}{\sqrt{50}}\\ & =& 1.697\\ & & \end{array}$

So, its shape will be symmetric and bell-shaped.

b.

Here the sample size is$n=50$ and the sample is drawn from the population having the mean$\mu =40$ and the standard deviation of $\sigma =12$.

So, the probability distribution of $\overline{x}$follows normal distribution with mean${\mu }_{\overline{x}}=40$ and standard deviation of ${\sigma }_{\overline{x}}=1.697$.

Therefore, the probability that a second random sample of size 50 would yield $\overline{x}$greater than or equal to 44 is,

$\begin{array}{rcl}\mathrm{Pr}\left(\overline{x}⩾44\right)& =& 1-\mathrm{Pr}\left(\overline{x}<44\right)\\ & =& 1-\mathrm{Pr}\left(\frac{\left(\overline{x}-{\mu }_{\overline{x}}\right)}{{\sigma }_{\overline{x}}}<\frac{44-40}{1.697}\right)\\ & =& 1-\mathrm{Pr}\left(z<2.35\right)\\ & =& 1-0.99061\\ & =& 0.0093\\ & & \end{array}$

Thus, the required probability is 0.0093.

c.

Consider the interval $\left(\mu \pm 2\frac{\sigma }{\sqrt{n}}\right)$. Where $\mu =40,\sigma =12 and n=50$.

Therefore, the interval is,

$\begin{array}{rcl}\left(\mu \pm 2\frac{\sigma }{\sqrt{n}}\right)& =& \left(40\pm 2\frac{12}{\sqrt{50}}\right)\\ & =& \left[\left(40-2\frac{12}{\sqrt{50}}\right),\left(40+2\frac{12}{\sqrt{50}}\right)\right]\\ & =& \left[\left(40-3.40\right),\left(40+3.40\right)\right]\\ & =& \left(36.60,43.40\right)\\ & & \end{array}$

Therefor, the probability that the mean will lies between this interval is,

$\begin{array}{rcl}\mathrm{Pr}\left(36.60⩽\overline{x}⩽43.39\right)& =& \mathrm{Pr}\left(\frac{36.60-40}{1.697}⩽\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}⩽\frac{43.40-40}{1.697}\right)\\ & =& \mathrm{Pr}\left(-2⩽z⩽2\right)\\ & =& \mathrm{Pr}\left(z⩽2\right)-\mathrm{Pr}\left(z⩽-2\right)\\ & =& 0.9772-0.0228\\ & =& 0.9544\\ & & \end{array}$

Thus, the required probability is 0.9544.