Chapter 4.4, Problem 42E

The average playing time of music albums in a large collection is 35 minutes, and the standard deviation is 5 minutes.

1. a. What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean? (Hint: See Example 4.14.)
2. b. Without assuming anything about the distribution of times, at least what percentage of the times are between 25 and 45 minutes? (Hint: See Example 4.15.)
3. c. Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than 20 minutes or greater than 50 minutes?
4. d. Assuming that the distribution of times is approximately normal, about what percentage of times are between 25 and 45 minutes? less than 20 minutes or greater than 50 minutes? less than 20 minutes?

To determine

Find the values that are 1 standard deviation above the mean and 1 standard deviation below the mean.

Obtain the values at 2 standard deviations away from the mean.

Answer to Problem 42E

The value at 1 standard deviation above the mean is 40 minutes.

The value at 1 standard deviation below the mean is 30 minutes.

The values at 2 standard deviations away from the mean are 25 minutes below and 45 minutes above the mean.

Explanation of Solution

Calculation:

The given information states that the mean playing time of the music albums in a large collection is 35 minutes with the corresponding standard deviation of 5 minutes.

The value at 1 standard deviation above the mean is obtained as follows:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 1 standard }\\ \text{deviation above mean}\end{array}\right\}=\text{mean}+\text{standard devition}\\ =35+5\\ =40\text{ minutes}\end{array}$

Thus, the value at 1 standard deviation above the mean is 40 minutes.

The value at 1 standard deviation below the mean is obtained as follows:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 1 standard }\\ \text{deviation below mean}\end{array}\right\}=\text{mean}-\text{standard devition}\\ =35-5\\ =30\text{ minutes}\end{array}$

Thus, the value at 1 standard deviation below the mean is 40 minutes.

The values at 2 standard deviations away from the mean are obtained as follows:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 2 standard deviation }\\ \text{away from the mean}\end{array}\right\}=\text{mean}\pm 2\text{standard devition}\\ =35\pm 2\left(5\right)\\ =35\pm 10\\ =\left(35-10,35+10\right)\\ =\left(25\text{ minutes},45\text{ minutes}\right)\end{array}$

Thus, the values at 2 standard deviations away from the mean are below 25 minutes and above 45 minutes

To determine

Find the least percentage of the times between 25 minutes and 45 minutes.

Answer to Problem 42E

The least percentage of the times between 25 minutes and 45 minutes is 75%.

Explanation of Solution

From the previous Part (a), it is clear that the values of 25 minutes and 45 minutes lie 2 standard deviations below and above the mean respectively.

Chebyshev’s rule:

For any number k, where $k\ge 1$, the percentage of the observation within k standard deviations of the mean is at least $100\left(1-\frac{1}{k^2}\right)\%$.

The z-score gives the number of standard deviations that an observation is away from the mean. Here, the z-scores for 25 minutes and 45 minutes are respectively –2 and 2. Thus, these observations of time are 2 standard deviations from the mean, that is, $k=2$.

Using Chebyshev’s rule,

$\begin{array}{c}100\left(1-\frac{1}{k^2}\right)\%=100\left(1-\frac{1}{2^2}\right)\%\\ =100\left(1-0.25\right)\%\\ =100\left(0.75\right)\%\\ =75\%\end{array}$

In this context, at least 0.75 proportion of observations lie within 2 standard deviations of the mean.

Therefore, the least percentage of the times between 25 minutes and 45 minutes is 75%.

To determine

Find the percentage of the times that are either less than 20 minutes or greater than 50 minutes.

Answer to Problem 42E

The percentage of the times that are either less than 20 minutes or greater than 50 minutes is 11%.

Explanation of Solution

Calculation:

The values at 3 standard deviations away the mean is obtained as follows:

$\begin{array}{c}\left\{\begin{array}{l}\text{value at 3 standard deviation }\\ \text{away from the mean}\end{array}\right\}=\text{mean}\pm 3\text{standard devition}\\ =35\pm 3\left(5\right)\\ =35\pm 15\\ =\left(35-15,35+15\right)\\ =\left(20\text{ minutes},50\text{ minutes}\right)\end{array}$

Therefore, the times that are either less than 20 minutes or greater than 50 minutes fall 3 standard deviations away from the mean.

Here, the z-scores for 20 minutes and 50 minutes are respectively –3 and 3. Thus, $k=3$.

Using Chebyshev’s rule,

$\begin{array}{c}100\left(\frac{1}{k^2}\right)\%=100\left(\frac{1}{3^2}\right)\%\\ =100\left(0.111\right)\%\\ \approx 11.1\%\end{array}$

Therefore, the percentage of the times that are either less than 20 minutes or greater than 50 minutes is 11%.

To determine

Find the percentage of times between 25 and 45 minutes for an approximately normal distribution.

Find the percentage of times less than 20 minutes or greater than 50 minutes for an approximately normal distribution.

Find the percentage of times less than 20 minutes for an approximately normal distribution.

Answer to Problem 42E

The percentage of times between 25 and 45 minutes for an approximately normal distribution is approximately 95%.

The percentage of times less than 20 minutes or greater than 50 minutes for an approximately normal distribution is approximately 0.3%.

The percentage of times less than 20 minutes for an approximately normal distribution is approximately 0.15%.

Explanation of Solution

Calculation:

Assume that the distribution of times is approximately normal.

Denote X as the observation.

By observing the previous Parts (a), (b), and, (c), the values at 25 and 45 minutes lie at 1 standard deviation from mean, the values at 20 and 50 minutes lie at 3 standard deviations from mean.

Empirical Rule:

The Empirical Rule for a Normal model states that:

• Within 1 standard deviation of mean, 68% of all observations will lie.
• Within 2 standard deviations of mean, 95% of all observations will lie.
• Within 3 standard deviations of mean, 99.7% of all observations will lie.

Based on the Empirical rule, 95% of the observation will fall between 2 standard deviations about mean.

Therefore, the percentage of times between 25 and 45 minutes for an approximately normal distribution is approximately 95%.

Based on the Empirical rule, 99.7% of the observation will fall between 1 standard deviation about mean.

The percentage of times less than 20 minutes and greater than 50 minutes is obtained as follows:

$\begin{array}{c}P\left(X<20\text{ or }X>50\right)=1-P\left(20<X<50\right)\\ =1-0.997\\ \approx 0.003\end{array}$

Therefore, the percentage of times less than 20 minutes and greater than 50 minutes for an approximately normal distribution is approximately 0.3%.

The percentage of times less than 20 minutes is obtained as follows:

$\begin{array}{c}P\left(X<20\right)=\frac{P\left(X<20\text{ or }X>50\right)}{2}\\ =\frac{0.003}{2}\\ =0.0015\end{array}$

Therefore, the percentage of times less than 20 minutes for an approximately normal distribution is approximately 0.15%.