Chapter 3.3, Problem 37E

Consider a sample with a mean of 30 and a standard deviation of 5. Use Chebyshev’s theorem to determine the percentage of the data within each of the following ranges:

1. a. 20 to 40
2. b. 15 to 45
3. c. 22 to 38
4. d. 18 to 42
5. e. 12 to 48

To determine

Find the percentage of the data within the range 20 to 40 using Chebyshev’s theorem.

Answer to Problem 37E

At least 75% of the data are within the range 20 to 40.

Explanation of Solution

Calculation:

The given information is that the mean and standard deviation of the sample are 30 and 5, respectively.

The formula for z-score is $z_i=\frac{x_i-\overline{x}}{s}$.

Where $\overline{x}$ is the sample mean and $s$ is the sample standard deviation.

The z-score corresponding to 20 can be obtained as follows:

Substitute $x_i=20$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{20-30}{5}\\ =-2\end{array}$

Thus, the z-score corresponding to 20 is –2.

The z-score corresponding to 40 can be obtained as follows:

Substitute $x_i=40$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{40-30}{5}\\ =2\end{array}$

Thus, the z-score corresponding to 40 is 2.

Chebyshev’s Theorem:

The z- standard deviations of the mean contain at least $\left(1-\frac{1}{z^2}\right)$ of the data, where z is any value greater than 1.

By applying the Chebychev’s theorem with $z=2$, the value of $\left(1-\frac{1}{z^2}\right)$ is

$\begin{array}{c}1-\frac{1}{2^2}=1-\frac{1}{4}\\ =\frac{4-1}{4}\\ =\frac{3}{4}\\ =0.75\end{array}$

Thus, at least 75% of the data are within the range 20 to 40.

To determine

Find the percentage of the data within the range 15 to 45, using Chebyshev’s theorem.

Answer to Problem 37E

At least 89% of the data are within the range 15 to 45.

Explanation of Solution

Calculation:

The z-score corresponding to 15 can be obtained as follows:

Substitute $x_i=15$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{15-30}{5}\\ =-3\end{array}$

Thus, the z-score corresponding to 15 is –3.

The z-score corresponding to 45 can be obtained as follows:

Substitute $x_i=45$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{45-30}{5}\\ =3\end{array}$

Thus, the z-score corresponding to 45 is 3.

By applying the Chebychev’s theorem with $z=3$, the value of $\left(1-\frac{1}{z^2}\right)$ is

$\begin{array}{c}1-\frac{1}{3^2}=1-\frac{1}{9}\\ =\frac{9-1}{9}\\ =\frac{8}{9}\\ =0.89\end{array}$

Thus, at least 89% of the data are within the range 15 to 45.

To determine

Find the percentage of the data within the range 22 to 38, using Chebyshev’s theorem.

Answer to Problem 37E

At least 61% of the data are within the range 22 to 38.

Explanation of Solution

Calculation:

The z-score corresponding to 22 can be obtained as follows:

Substitute $x_i=22$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{22-30}{5}\\ =-1.6\end{array}$

Thus, the z-score corresponding to 22 is –1.6.

The z-score corresponding to 38 can be obtained as follows:

Substitute $x_i=38$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{38-30}{5}\\ =1.6\end{array}$

Thus, the z-score corresponding to 38 is 1.6.

By applying the Chebychev’s theorem with $z=1.6$, the value of $\left(1-\frac{1}{z^2}\right)$ is

$\begin{array}{c}1-\frac{1}{{1.6}^2}=1-\frac{1}{2.56}\\ =\frac{2.56-1}{2.56}\\ =\frac{1.56}{2.56}\\ =0.61\end{array}$

Thus, at least 61% of the data are within the range 22 to 38.

To determine

Find the percentage of the data within the range 18 to 42, using Chebyshev’s theorem.

Answer to Problem 37E

At least 83% of the data are within the range 18 to 42.

Explanation of Solution

Calculation:

The z-score corresponding to 18 can be obtained as follows:

Substitute $x_i=18$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{18-30}{5}\\ =-2.4\end{array}$

Thus, the z-score corresponding to 18 is –2.4.

The z-score corresponding to 42 can be obtained as follows:

Substitute $x_i=42$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{42-30}{5}\\ =2.4\end{array}$

Thus, the z-score corresponding to 42 is 2.4.

By applying the Chebychev’s theorem with $z=2.4$, the value of $\left(1-\frac{1}{z^2}\right)$ is

$\begin{array}{c}1-\frac{1}{{2.4}^2}=1-\frac{1}{5.76}\\ =\frac{5.76-1}{5.76}\\ =\frac{4.76}{5.76}\\ =0.83\end{array}$

Thus, at least 83% of the data are within the range 18 to 42.

To determine

Find the percentage of the data within the range 12 to 48, using Chebyshev’s theorem.

Answer to Problem 37E

At least 92% of the data are within the range 12 to 48.

Explanation of Solution

Calculation:

The z-score corresponding to 12 can be obtained as follows:

Substitute $x_i=12$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{12-30}{5}\\ =\frac{-18}{5}\\ =-3.6\end{array}$

Thus, the z-score corresponding to 12 is –3.6.

The z-score corresponding to 48 can be obtained as follows:

Substitute $x_i=48$, $s=30$ and $\overline{x}=5$

$\begin{array}{c}z=\frac{48-30}{5}\\ =\frac{18}{5}\\ =3.6\end{array}$

Thus, the z-score corresponding to 48 is 3.6.

By applying the Chebychev’s theorem with $z=3.6$, the value of $\left(1-\frac{1}{z^2}\right)$ is

$\begin{array}{c}1-\frac{1}{{3.6}^2}=1-\frac{1}{12.96}\\ =\frac{12.96-1}{12.96}\\ =\frac{11.96}{12.96}\\ =0.92\end{array}$

Thus, at least 92% of the data are within the range 12 to 48.