Question

Checkout lanes at a supermarket. A team of consultants working for a large national supermarket chain based in the New York metropolitan area developed a statistical model for predicting the annual sales of potential new store locations. Part of their analysis involved identifying variables that influence store sales, such as the size of the store (in square feet), the size of the surrounding population, and the number of checkout lanes. They surveyed 52 supermarkets in a particular region of the country and constructed the relative frequency distribution shown below to describe the number of checkout lanes per store, x.

a. Why do the relative frequencies in the table represent the approximate probabilities of a randomly selected supermarket having x number of checkout lanes?

b. Find$\mathit{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and interpret its value in the context of the problem.

c. Find the standard deviation of x.

d. According to Chebyshev’s Rule (Chapter 2, p. 106), what percentage of supermarkets would be expected to fall within$\mathit{\mu }\mathbf{\pm }\mathit{\sigma }$? within$\mathit{\mu }\mathbf{\pm }\mathbf{2}\mathit{\sigma }$?

e. What is the actual number of supermarkets that fall within? ? Compare your answers with those of part d. Are the answers consistent?

Short answer

a. it is observed that total frequency all relative frequencies are greater than are equal to zero and total probability is equal to one.

b.On average the number of checkout lanes per store will equal to 6.5.

c. The standard deviation of x is 1.9974.

d. The percentage of supermarket would be expected to fall within $\mu \pm \sigma$is 68 and the percentage of supermarket would be expected to fall within $\mu \pm 2\sigma$is 95%.

e. The actual number of supermarkets that fall within$\mu \pm 2\sigma$ is 47.27 and the actual number of supermarkets that fall within $\mu \pm 2\sigma$is 94.57.By comparing the answers to those parts (d) and part (c) the answers are not consistent.

Step-by-step solution

Given information

A team of consultants surveyed 52 supermarkets in a particular region of the country and constructed the relative frequency distribution.

Step 2:

a.

Since, the relative frequency is the ratio of the actual frequency and total frequency. The relative frequency itself called as probability distribution.

So, the relative frequencies in the above table represent the approximate probabilities of randomly supermarket having x number of checkout lanes. From the above relative frequency distribution, it is observed that total frequency all relative frequencies are greater than are equal to zero and total probability is equal to one.

Calculating the mean of x 

b.

Since,

$\begin{array}{c}\mathbf{\mu }\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ \mathbf{=}\mathbf{\sum }\mathbf{x}\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\end{array}$

$\begin{array}{c}=1\left(0.01\right)+2\left(0.04\right)+3\left(0.04\right)+4\left(0.08\right)+5\left(0.10\right)+6\left(0.15\right)\\ +7\left(0.25\right)+8\left(0.20\right)+9\left(0.08\right)+10\left(0.05\right)\\ =6.5\end{array}$

On average the number of checkout lanes per store will equal to 6.5.

Calculating the standard deviation of x 

c.

Let,

${\mathit{\sigma }}^{\mathbf{2}}\mathbf{=}\mathit{E}\mathbf{\left[}{\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{\mu }\mathbf{)}}^{\mathbf{2}}\mathbf{\right]}$

$\begin{array}{l}=\sum {(x-\mu )}^{2}p\left(x\right)\\ ={(1-6.5)}^2\left(0.01\right)+{(2-6.5)}^2\left(0.04\right)+{(3-6.5)}^2\left(0.04\right)+\mathrm{...}+{(10-6.5)}^2\left(0.05\right)\\ =3.99\end{array}$

localid="1658223087292" $\begin{array}{c}\sigma =\sqrt{{\sigma }^2}\\ =\sqrt{3.99}\\ =1.9974\end{array}$

Therefore, the standard deviation of x is 1.9974.

Calculating the Percentage

d.

According to Chebyshev’s rule, the percentage of supermarket would be expected to fall within$\mu \pm \sigma$ is 68. The percentage of supermarket would be expected to fall within$\mu \pm 2\sigma$ is 95%.

Calculating the actual number of supermarket 

e.

The actual number of supermarkets that fall within $\mu \pm 2\sigma$is shown below:

role="math" localid="1658222683820" $\begin{array}{c}(\mu \pm 2\sigma )=(\mu -\sigma ,\mu +\sigma )\\ =(6.5-1.9974,6.5-1.9974)\\ =(4.5026,8.4974)\end{array}$

Now, the actual probability that x falls in the interval includes the sum of the values is $5+6+7+8=26$

Total of the all x values=55

The percentage

$\begin{array}{l}=\frac{26}{55}\times 100\\ =47.27\end{array}$

Hence, the actual number of supermarkets that fall within$\mu \pm 2\sigma$is 47.27

Also, the actual number of supermarkets that fall within $\mu \pm 2\sigma$is shown below:

$\begin{array}{c}\mu \pm 2\sigma =(\mu -2\sigma ,\mu +2\sigma )\\ =(6.5-2\left(1.9974\right),6.5-2\left(1.9974\right))\\ =(2.5052,10.4948)\end{array}$

The actual probability that x falls in the interval includes the sum of the values is

$3+4+5+6+7+8+9+10=52$

Total of the all xvalues=55

The percentage

$\begin{array}{l}=\frac{52}{55}\times 100\\ =94.57\end{array}$

Hence, the actual number of supermarkets that fall within i$\mu \pm 2\sigma$s 94.57.

By comparing the answers to those parts (d) and part (c) the answers are not consistent.