91Ó°ÊÓ

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

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.

b.

Since,

$\begin{array}{c}\mathbf{μ}\mathbf{=}\mathbf{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ \mathbf{=}\mathbf{∑}\mathbf{x}\mathbf{\text{ }}\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\end{array}$

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

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

c.

Let,

${\mathit{σ}}^{\mathbf{2}}\mathbf{=}\mathit{E}\mathbf{\left[}{\mathbf{(}\mathbf{x}\mathbf{−}\mathbf{μ}\mathbf{)}}^{\mathbf{2}}\mathbf{\right]}$

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

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

Therefore, the standard deviation of x is 1.9974.

d.

According to Chebyshev’s rule, the percentage of supermarket would be expected to fall within$\mathrm{μ}Â\pm \mathrm{σ}$ is 68. The percentage of supermarket would be expected to fall within$\mathrm{μ}Â\pm 2\mathrm{σ}$ is 95%.

e.

The actual number of supermarkets that fall within $\mathrm{μ}Â\pm 2\mathrm{σ}$is shown below:

role="math" localid="1658222683820" $\begin{array}{c}(\mathrm{μ}Â\pm 2\mathrm{σ})=(\mathrm{μ}−\mathrm{σ},\mathrm{μ}+\mathrm{σ})\\ =(6.5−1.9974,6.5−1.9974)\\ =(4.5026,8.4974)\end{array}$

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

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

Hence, the actual number of supermarkets that fall within$\mathrm{μ}Â\pm 2\mathrm{σ}$is 47.27

Also, the actual number of supermarkets that fall within $\mathrm{μ}Â\pm 2\mathrm{σ}$is shown below:

$\begin{array}{c}\mathrm{μ}Â\pm 2\mathrm{σ}=(\mathrm{μ}−2\mathrm{σ},\mathrm{μ}+2\mathrm{σ})\\ =(6.5−2\left(1.9974\right),6.5−2\left(1.9974\right))\\ =(2.5052,10.4948)\end{array}$

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

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

Hence, the actual number of supermarkets that fall within i$\mathrm{μ}Â\pm 2\mathrm{σ}$s 94.57.

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