Questions: A cellular phone company monitors monthly phone usage. The following data represent the monthly phone use in minutes of one particular customer for the past 20 months. Use the given data to answer parts (a) and (b). 324 475 538 544 453 468 393 377 433 405 330 388 511 536 365 461 500 513 415 474 (a) Determine the standard deviation and interquartile range of the data. s= (Round to two decimal places as needed.) IQR= (Type an integer or a decimal.) (b) Suppose the month in which the customer used 324 minutes was not actually that customer's phone. That particular month the customer did not use their phone at all, so 0 minutes were used. How does changing the observation from 324 to 0 affect the standard deviation and interquartile range? What property does this illustrate? The standard deviation and the interquartile range What property does this illustrate? Choose the correct answer below. Empirical Rule Weighted Mean Dispersion Resistance

Solution Steps

The mean \( \mu \) of the monthly phone usage data is calculated as follows:

\[ \mu = \frac{\sum x_i}{n} = \frac{8903}{20} = 445.15 \]

Next, the variance \( \sigma^2 \) is computed using the formula:

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 4674.34 \]

The standard deviation \( \sigma \) is then obtained by taking the square root of the variance:

\[ \sigma = \sqrt{4674.34} = 68.37 \]

To find the interquartile range, we first sort the data:

\[ \text{Sorted Data} = [324, 330, 365, 377, 388, 393, 405, 415, 433, 453, 461, 468, 474, 475, 500, 511, 513, 536, 538, 544] \]

Next, we calculate the first quartile \( Q_1 \) and the third quartile \( Q_3 \):

For \( Q_1 \):

\[ \text{Rank} = 0.25 \times (20 + 1) = 5.25 \] \[ Q_1 = \frac{X_{\text{lower}} + X_{\text{upper}}}{2} = \frac{388 + 393}{2} = 390.5 \]

For \( Q_3 \):

\[ \text{Rank} = 0.75 \times (20 + 1) = 15.75 \] \[ Q_3 = \frac{X_{\text{lower}} + X_{\text{upper}}}{2} = \frac{500 + 511}{2} = 505.5 \]

The interquartile range \( \text{IQR} \) is then calculated as:

\[ \text{IQR} = Q_3 - Q_1 = 505.5 - 390.5 = 115.0 \]

If we change the observation from 324 to 0, the new dataset becomes:

\[ \text{Modified Data} = [0, 330, 365, 377, 388, 393, 405, 415, 433, 453, 461, 468, 474, 475, 500, 511, 513, 536, 538, 544] \]

We recalculate the mean \( \mu \):

\[ \mu = \frac{\sum x_i}{n} = \frac{8579}{20} = 428.95 \]

Next, we compute the new variance \( \sigma^2 \):

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 14055.0 \]

The modified standard deviation \( \sigma \) is:

\[ \sigma = \sqrt{14055.0} = 118.55 \]

The sorted modified data is:

\[ \text{Sorted Modified Data} = [0, 330, 365, 377, 388, 393, 405, 415, 433, 453, 461, 468, 474, 475, 500, 511, 513, 536, 538, 544] \]

We recalculate \( Q_1 \) and \( Q_3 \):

For \( Q_1 \):

\[ \text{Rank} = 0.25 \times (20 + 1) = 5.25 \] \[ Q_1 = \frac{388 + 393}{2} = 390.5 \]

For \( Q_3 \):

\[ \text{Rank} = 0.75 \times (20 + 1) = 15.75 \] \[ Q_3 = \frac{500 + 511}{2} = 505.5 \]

The modified interquartile range \( \text{IQR} \) remains:

\[ \text{IQR} = Q_3 - Q_1 = 505.5 - 390.5 = 115.0 \]

The change in standard deviation from \( 68.37 \) to \( 118.55 \) and the unchanged IQR of \( 115.0 \) illustrates the concept of resistance. The IQR is resistant to outliers, while the standard deviation is not.

Final Answer