Two groups of students, Group A and Group B, have the age distributions shown below. Which statement about the distributions is true?

Question

There are two histograms titled distribution A and the other as distribution B, both having age on the horizontal axis as 15, 16, 17, 18, and 19 and number of students from 0 to 7 along the vertical axis. Histogram titled distribution A shows, the bar for 15 at 2, the bar for 16 at 5, the bar for 17 at 6, the bar for 18 at 5, and the bar for 19 at 2. Histogram titled distribution B shows, the bar for 15 at 2, the bar for 16 at 3, the bar for 17 at 4, the bar for 18 at 5, and the bar for 19 at 6.
A.
Distribution B has a larger mean, but the medians of both distributions are the same.

B.
Distribution B has both a larger mean and a larger median than distribution A.

C.
Distribution B has a smaller mean, but the medians of both distributions are the same.

D.
Distribution B has both a smaller mean and a smaller median than distribution A.

Answer 1

To evaluate the mean and median of each distribution, let's first summarize the data given for both distributions:

Distribution A:

• Ages:
  • 15 → 2 students
  • 16 → 5 students
  • 17 → 6 students
  • 18 → 5 students
  • 19 → 2 students

Distribution B:

• Ages:
  • 15 → 2 students
  • 16 → 3 students
  • 17 → 4 students
  • 18 → 5 students
  • 19 → 6 students

Calculate the Mean

For the mean of both distributions, we will use the formula \(\text{Mean} = \frac{\sum (x \cdot f)}{N}\), where \(x\) is the age and \(f\) is the frequency (number of students) for that age, and \(N\) is the total number of students.

Mean of Distribution A:

• Total number of students \(N_A = 2 + 5 + 6 + 5 + 2 = 20\)
• Mean: \[ \text{Mean}_A = \frac{(15 \cdot 2) + (16 \cdot 5) + (17 \cdot 6) + (18 \cdot 5) + (19 \cdot 2)}{20} = \frac{30 + 80 + 102 + 90 + 38}{20} = \frac{340}{20} = 17 \]

Mean of Distribution B:

• Total number of students \(N_B = 2 + 3 + 4 + 5 + 6 = 20\)
• Mean: \[ \text{Mean}_B = \frac{(15 \cdot 2) + (16 \cdot 3) + (17 \cdot 4) + (18 \cdot 5) + (19 \cdot 6)}{20} = \frac{30 + 48 + 68 + 90 + 114}{20} = \frac{350}{20} = 17.5 \]

Calculate the Median

To find the median, we need to arrange the frequency data and find the middle value:

• For both distributions with \(N = 20\), the median will be the average of the 10th and 11th values.

Distribution A:

• Ordered ages:
  • 2 (Age 15), 5 (Age 16), 6 (Age 17), 5 (Age 18), 2 (Age 19)
• Cumulative counts:
  • Age 15: 2
  • Age 16: 7 (2+5)
  • Age 17: 13 (7+6)
  • Age 18: 18 (13+5)
  • Age 19: 20 (18+2)

The 10th and 11th values are both from Age 17.

Median of Distribution A: \[ \text{Median}_A = 17 \]

Distribution B:

• Ordered ages:
  • 2 (Age 15), 3 (Age 16), 4 (Age 17), 5 (Age 18), 6 (Age 19)
• Cumulative counts:
  • Age 15: 2
  • Age 16: 5 (2+3)
  • Age 17: 9 (5+4)
  • Age 18: 14 (9+5)
  • Age 19: 20 (14+6)

The 10th and 11th values are from Age 18.

Median of Distribution B: \[ \text{Median}_B = 18 \]

Summary:

• Mean of Distribution A: \(17\), Mean of Distribution B: \(17.5\) → Distribution B has a larger mean.
• Median of Distribution A: \(17\), Median of Distribution B: \(18\) → Distribution B has a larger median.

Conclusion:

The statements can now be compared to the computed means and medians:

• A. Distribution B has a larger mean, but the medians of both distributions are the same. (Incorrect)
• B. Distribution B has both a larger mean and a larger median than distribution A. (Correct)
• C. Distribution B has a smaller mean, but the medians of both distributions are the same. (Incorrect)
• D. Distribution B has both a smaller mean and a smaller median than distribution A. (Incorrect)

Thus, the correct answer is B.