Question

dillon and maura receive emails from clients. the table shows the numbers of emails each employee received on six randomly selected days. based on the data in the table, which statement best describes the number of emails from clients on a typical day? on a typical day, dillon receives more emails from clients than maura because the mean of dillons data is greater than the mean of mauras data. on a typical day, dillon receives more emails from clients than maura because the range of dillons data is greater than the range of mauras data. on a typical day, maura receives more emails from clients than dillon because the mean of mauras data is greater than the mean of dillons data. on a typical day, maura receives more emails from clients than dillon because the range of mauras data is greater than the range of dillons data.

Explanation:

Step1: Calculate the mean of Dillon's data

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). For Dillon's data \(x_{1}=34,x_{2}=26,x_{3}=31,x_{4}=25,x_{5}=23,x_{6}=40\), \(n = 6\).
\(\sum_{i=1}^{6}x_{i}=34 + 26+31+25+23+40=179\)
\(\bar{x}_{Dillon}=\frac{179}{6}\approx29.83\)

Step2: Calculate the mean of Maura's data

For Maura's data \(x_{1}=28,x_{2}=41,x_{3}=38,x_{4}=26,x_{5}=44,x_{6}=30\), \(n = 6\)
\(\sum_{i = 1}^{6}x_{i}=28+41+38+26+44+30 = 207\)
\(\bar{x}_{Maura}=\frac{207}{6}=34.5\)

Step3: Calculate the range of Dillon's data

The range \(R=\text{Max}-\text{Min}\). For Dillon's data, \(\text{Max}=40\), \(\text{Min}=23\), \(R_{Dillon}=40 - 23=17\)

Step4: Calculate the range of Maura's data

For Maura's data, \(\text{Max}=44\), \(\text{Min}=26\), \(R_{Maura}=44 - 26 = 18\)

Since we use the mean to describe the central - tendency (a typical day) and \(34.5>29.83\) (mean of Maura's data \(>\) mean of Dillon's data)

Answer:

C. On a typical day, Maura receives more emails from clients than Dillon because the mean of Maura's data is greater than the mean of Dillon's data.