Question

the tuition and fees (in thousands of dollars) for the top 14 universities in a recent year are listed below. find the mean, median, and mode of the data, if possible. if any of these measures cannot be found or a measure does not represent the center of the data, explain why.
40 46 39 44 45 35 40
44 44 45 47 44 44 40

Explanation:

Step1: Find the Mean

First, sum all the data points. The data is: 40, 46, 39, 44, 45, 35, 40, 44, 44, 45, 47, 44, 44, 40.
Sum = \(35 + 39 + 40 + 40 + 40 + 44 + 44 + 44 + 44 + 44 + 45 + 45 + 46 + 47\)
Calculate each part: \(35+39 = 74\); \(40\times3 = 120\); \(44\times5 = 220\); \(45\times2 = 90\); \(46 + 47 = 93\)
Total sum = \(74+120 + 220+90 + 93= 697\)
Mean = \(\frac{\text{Sum}}{\text{Number of data points}}=\frac{697}{14}\approx49.79\)? Wait, no, wait, 14 data points? Wait, let's count again. The first row: 40,46,39,44,45,35,40 (7 numbers). Second row:44,44,45,47,44,44,40 (7 numbers). Total 14. Wait, but 35 + 39 is 74, 40+40+40=120 (total 74+120=194), 44+44+44+44+44=220 (194+220=414), 45+45=90 (414+90=504), 46+47=93 (504+93=597). Oh! I made a mistake earlier. So sum is 597. Then mean is \(\frac{597}{14}\approx42.64\) (wait, 14*42=588, 597-588=9, so 42 + 9/14 ≈42.64).

Step2: Find the Median

First, order the data from least to greatest:
35, 39, 40, 40, 40, 44, 44, 44, 44, 44, 45, 45, 46, 47.
There are 14 data points, so the median is the average of the 7th and 8th values.
7th value: 44, 8th value: 44.
Median = \(\frac{44 + 44}{2}=44\).

Step3: Find the Mode

The mode is the value that appears most frequently.
Count the frequency of each number:

• 35: 1
• 39: 1
• 40: 3
• 44: 5
• 45: 2
• 46: 1
• 47: 1

So 44 appears most often (5 times), so mode is 44.

Answer:

Mean: \(\approx 42.64\) (thousand dollars), Median: \(44\) (thousand dollars), Mode: \(44\) (thousand dollars)