Consider the following data set.

a. Draw a graph similar to Fig. $3.5$on page $111$.

b. Compare the percentage of the observations that actually lie within two standard deviations to either side of the mean with that given by Chebyshev's rule with localid="1651240279191" $k=2$

c. Repeat part (b) with localid="1651240341435" $k=3$

We need to plot the graph of the given data

We know that,

Chebyshev's rule is as follows,

$100\left(1-\frac{1}{k^2}\right)\%$of the given data where $k$is the standard deviation from mean.

The mean is calculated as the sum of all values divided by no. of values,

As a result, mean is,

$\overline{x}=\frac{30+16+22+23+18+18+20+24+19+13+9+28}{12}=20$

The standard deviation is the square root of variance and variance is the sum of squared deviations divided by the number of values$-1$.

$s=\sqrt{\frac{{\left(30-20\right)}^2+{\left(16-20\right)}^2+....+{\left(9-20\right)}^2+{\left(58-20\right)}^2}{12-1}}=5.9$

Therefore,

$$\begin{gathered} \overline{x}-s=20-5.9=14.1 \\ \overline{x}+s=20+5.9=25.9 \\ \overline{x}-2s=20-\left(2\times 5.9\right)=8.2 \\ \overline{x}+2s=20+\left(2\times 5.9\right)=31.8 \\ \overline{x}-3s=20-\left(3\times 5.9\right)=2.3 \\ \overline{x}+3s=20+\left(3\times 5.9\right)=37.7 \end{gathered}$$

The graph is as follows,

We are given that$k=2$

We know that,

Chebyshev's rule is given as,

$100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{2^2}\right)\%=100\left(1-\frac{1}{4}\right)\%=100\left(\frac{3}{4}\right)\%=75\%$

It is within the $2$standard deviations from the mean.

We note in the graph that mainly deviations lie between $\overline{x}-2s$and $\overline{x}+2s$

Therefore, actual proportion is, $100\%$

Which is a lot higher than Chebyshev's rule.

We are given that$k=3$

By Chebyshev's rule,

$100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{3^2}\right)\%=100\left(\frac{8}{9}\right)\%=89\%$

It is within the $3$standard deviations from the mean.

We note in the graph that mainly deviations lie between $\overline{x}-3s$and$\overline{x}+3s$

Therefore, actual proportion is,$100\%$

Which is a lot higher than Chebyshev's rule.