UpStudy Free Solution:

To determine which applicant has the highest mean and median credit scores, we need to calculate both the mean and the median for each client.

First, let's calculate the mean and median for each client:

1. Neil:

- Scores: 726, 752, 822

- Mean: \(\frac { 726 + 752 + 822} { 3} = \frac { 2300} { 3} \approx 766.67\)

- Median: 752 (the middle value when scores are ordered)

2. Paula:

- Scores: 634, 732, 771

- Mean: \(\frac { 634 + 732 + 771} { 3} = \frac { 2137} { 3} \approx 712.33\)

- Median: 732 (the middle value when scores are ordered)

3. Jeff:

- Scores: 721, 760, 754

- Mean: \(\frac { 721 + 760 + 754} { 3} = \frac { 2235} { 3} \approx 745.00\)

- Median: 754 (the middle value when scores are ordered)

4. Peter:

- Scores: 666, 716, 707

- Mean: \(\frac { 666 + 716 + 707} { 3} = \frac { 2089} { 3} \approx 696.33\)

- Median: 707 (the middle value when scores are ordered)

Now, let's compare the means and medians:

- Mean scores:

- Neil: 766.67

- Paula: 712.33

- Jeff: 745.00

- Peter: 696.33

- Median scores:

- Neil: 752

- Paula: 732

- Jeff: 754

- Peter: 707

From these calculations:

- Neil has the highest mean score (766.67).

- Jeff has the highest median score (754).

Therefore, the correct answer is:

b. Neil has the highest mean score, but Jeff has the highest median score.

Supplemental Knowledge:

In statistics, the mean (or average) is calculated by summing all values and dividing by the number of values. The median is the middle value when the data points are arranged in ascending order. If there is an even number of observations, the median is the average of the two middle numbers.

For example, to find the mean and median of a set of scores:

1. Mean:

\[\text { Mean} = \frac { \sum \text { Scores} } { \text { Number of Scores} } \]

2. Median:

- Arrange scores in ascending order.

- If odd number of scores, median is the middle score.

- If even number of scores, median is the average of the two middle scores.

Join millions who trust UpStudy with their academic journey; each challenge becomes an opportunity to expand!