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. 324 412 534 368 476 538 320 490 482 508 450 529 374 472 511 537 540 465 485 333 (a) Determine the standard deviation and interquartile range of the data. s= (Round to two decimal places as needed.)

Solution Steps

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

\[ \mu = \frac{\sum x_i}{n} = \frac{9148}{20} = 457.4 \]

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

\[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 5702.25 \]

The standard deviation \( \sigma \) is the square root of the variance:

\[ \sigma = \sqrt{5702.25} = 75.51 \]

To find \( Q1 \), we first sort the data:

\[ \text{Sorted data} = [320, 324, 333, 368, 374, 412, 450, 465, 472, 476, 482, 485, 490, 508, 511, 529, 534, 537, 538, 540] \]

Using the formula for the rank:

\[ \text{Rank} = Q \times (N + 1) = 0.25 \times (20 + 1) = 5.25 \]

We find \( Q1 \) by averaging the values at ranks 5 and 6:

\[ Q1 = \frac{X_{\text{lower}} + X_{\text{upper}}}{2} = \frac{374 + 412}{2} = 393.0 \]

Using the same sorted data, we calculate \( Q3 \):

\[ \text{Rank} = Q \times (N + 1) = 0.75 \times (20 + 1) = 15.75 \]

We find \( Q3 \) by averaging the values at ranks 15 and 16:

\[ Q3 = \frac{X_{\text{lower}} + X_{\text{upper}}}{2} = \frac{511 + 529}{2} = 520.0 \]

The interquartile range is calculated as:

\[ \text{IQR} = Q3 - Q1 = 520.0 - 393.0 = 127.0 \]

Final Answer