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The dot plots represent the cost of first-year college tuitions for two different U.S. regions. Which of the statements are true?

Dot plot display showing the cost of first year tuition in region 1 and 2, in US $ in thousands. Readings show number of universities and cost, in region 1: 2, $5; 1, $15; 3, $17; 2, $19, 1, $37. In region 2: 2, $5; 3, $8; 4, $9; 3, $31; 1, $37. 

I. Region 1 has a greater mean than region 2.
II. Region 2 has a greater median than region 1.
III. Region 1 has a greater interquartile range than region 2.
IV. Region 2 has a greater standard deviation than region 1.

A.
III and IV
B.
I and III
C.
II and IV
D.
I and II

1 answer

To evaluate the statements, we need to calculate the mean, median, interquartile range (IQR), and standard deviation for both regions based on the tuition costs given in the dot plots.

Region 1 costs (in thousands):

  • $5: 2 universities
  • $15: 1 university
  • $17: 3 universities
  • $19: 2 universities
  • $37: 1 university

This gives us the following list of data points for Region 1: \[ [5, 5, 15, 17, 17, 17, 19, 19, 37] \]

  1. Mean of Region 1: \[ \text{Mean} = \frac{(5 + 5 + 15 + 17 + 17 + 17 + 19 + 19 + 37)}{9} = \frac{ 5 + 5 + 15 + 17 + 17 + 17 + 19 + 19 + 37 }{9} = \frac{ 9 + 15 + 107 }{9} = \frac{122}{9} \approx 13.56 \]

  2. Median of Region 1: Since there are 9 data points, the median is the 5th value: \[ \text{Median} = 17 \]

  3. IQR of Region 1: The first quartile (Q1) is the median of the first half (first 4 data points): Q1 = (5 + 15) / 2 = 10 The third quartile (Q3) is the median of the second half (last 4 data points): Q3 = (17 + 19) / 2 = 18 \[ \text{IQR} = Q3 - Q1 = 18 - 10 = 8 \]

  4. Standard Deviation of Region 1: \[ \text{Variance} = \frac{(5-13.56)^2 + (5-13.56)^2 + (15-13.56)^2 + (17-13.56)^2 + (17-13.56)^2 + (17-13.56)^2 + (19-13.56)^2 + (19-13.56)^2 + (37-13.56)^2}{9} \approx \frac{81.06 + 81.06 + 1.97 + 11.66 + 11.66 + 11.66 + 29.58 + 29.58 + 552.87}{9} = \text{Variance} \approx 66.92 \text{Standard Deviation} \approx 8.19 \]

Region 2 costs (in thousands):

  • $5: 2 universities
  • $8: 3 universities
  • $9: 4 universities
  • $31: 3 universities
  • $37: 1 university

This gives us the following list of data points for Region 2: \[ [5, 5, 8, 8, 8, 9, 9, 9, 9, 31, 31, 31, 37] \]

  1. Mean of Region 2: \[ \text{Mean} = \frac{(5 + 5 + 8 + 8 + 8 + 9 + 9 + 9 + 9 + 31 + 31 + 31 + 37)}{13} = \frac{ 5 + 5 + 24 + 36 + 31 + 37 }{13} = \frac{ 5 + 5 + 24 + 36 + 31 + 37 }{13} = \frac{138}{13} \approx 10.62 \]

  2. Median of Region 2: Since there are 13 data points, the median is the 7th value: \[ \text{Median} = 9 \]

  3. IQR of Region 2: The first quartile (Q1) is the median of the first half (first 6 data points): Q1 = (5 + 8) / 2 = 6.5 The third quartile (Q3) is the median of the second half (last 6 data points): Q3 = (31 + 31) / 2 = 31 \[ \text{IQR} = Q3 - Q1 = 31 - 6.5 = 24.5 \]

  4. Standard Deviation of Region 2: \[ \text{Variance}\approx \frac{(5-10.62)^2 + (5-10.62)^2 + (8-10.62)^2 + (8-10.62)^2 + (8-10.62)^2 + (9-10.62)^2 + (9-10.62)^2 + (9-10.62)^2 + (9-10.62)^2 + (31-10.62)^2 + (31-10.62)^2 + (31-10.62)^2 + (37-10.62)^2}{13} \] This value will be larger than that of Region 1 since there are more data points far from the mean in Region 2.

Now, let's analyze the statements:

I. Region 1 has a greater mean than region 2.
False, because about 13.56 > 10.62.

II. Region 2 has a greater median than region 1.
False, because 9 < 17.

III. Region 1 has a greater interquartile range than region 2.
False, 8 < 24.5.

IV. Region 2 has a greater standard deviation than region 1.
True, as indicated by both higher data dispersion and swift calculation suggestions.

Final Answer:
A. III and IV