Question

the table lists the heights (in centimeters) of preschool girls and boys on a playground.
heights of preschool boys (centimeters) | heights of preschool girls (centimeters)
105.1 | 104.8
101.3 | 87
86.7 | 95
93.8 | 92.1
92.4 | 100
85.2 | 90.3
99.6 | 98.6
97.5 | 101.7
102.9 | 89.4
107 | 92.1
based on the table, which of the following is true?
a. the difference of the medians is about one - half the interquartile range of either data set.
b. the difference of the medians is about one - fourth the interquartile range of either data set.
c. the medians cannot be compared based on their interquartile ranges because the interquartile ranges are 1 centimeter apart.
d. the medians cannot be compared based on their interquartile ranges because the interquartile ranges are 3 centimeters apart.

Explanation:

Step1: Organize the data for boys and girls

First, we need to sort the data for preschool boys and preschool girls.

For Preschool Boys (heights in cm): \( 85.2, 86.7, 92.4, 93.8, 97.5, 99.6, 101.3, 102.9, 105.1, 107 \) Wait, wait, actually, let's count the number of data points. Wait, the table has 10 entries for boys? Wait, no, let's check the table again. Wait, the first column: 105.1, 101.3, 86.7, 93.8, 92.4, 85.2, 99.6, 97.5, 102.9, 107. Wait, that's 10 data points? Wait, no, let's count: 105.1 (1), 101.3 (2), 86.7 (3), 93.8 (4), 92.4 (5), 85.2 (6), 99.6 (7), 97.5 (8), 102.9 (9), 107 (10). Wait, no, maybe I miscounted. Wait, the table for boys: let's list all the values:

Boys: 105.1, 101.3, 86.7, 93.8, 92.4, 85.2, 99.6, 97.5, 102.9, 107. Wait, that's 10 values? Wait, no, maybe I made a mistake. Wait, the original table: let's check the number of rows. The table has 10 rows (from 105.1 to 107 for boys, and same for girls). Wait, no, let's count the rows:

Row 1: 105.1 (boys), 104.8 (girls)

Row 2: 101.3, 87

Row 3: 86.7, 95

Row 4: 93.8, 92.1

Row 5: 92.4, 100

Row 6: 85.2, 90.3

Row 7: 99.6, 98.6

Row 8: 97.5, 101.7

Row 9: 102.9, 89.4

Row 10: 107, 92.1

So both boys and girls have 10 data points (n = 10).

Step2: Find the median for boys and girls

For a data set with n = 10 (even number), the median is the average of the 5th and 6th values when sorted.

First, sort the boys' data:

Boys' heights: 85.2, 86.7, 92.4, 93.8, 97.5, 99.6, 101.3, 102.9, 105.1, 107. Wait, no, wait, let's sort them in ascending order:

85.2, 86.7, 92.4, 93.8, 97.5, 99.6, 101.3, 102.9, 105.1, 107. Wait, 85.2 (1), 86.7 (2), 92.4 (3), 93.8 (4), 97.5 (5), 99.6 (6), 101.3 (7), 102.9 (8), 105.1 (9), 107 (10). So the 5th value is 97.5, 6th is 99.6. So median for boys: \( \frac{97.5 + 99.6}{2} = \frac{197.1}{2} = 98.55 \) cm.

Now girls' heights: let's sort them:

87, 89.4, 90.3, 92.1, 92.1, 95, 98.6, 100, 101.7, 104.8. Wait, let's list all girls' heights: 104.8, 87, 95, 92.1, 100, 90.3, 98.6, 101.7, 89.4, 92.1. Now sort them in ascending order:

87, 89.4, 90.3, 92.1, 92.1, 95, 98.6, 100, 101.7, 104.8.

So 10 data points, median is average of 5th and 6th values. 5th value: 92.1, 6th value: 95. So median for girls: \( \frac{92.1 + 95}{2} = \frac{187.1}{2} = 93.55 \) cm. Wait, no, wait, 5th and 6th: positions 5 and 6 (since n=10, positions 1-10). So sorted girls' data:

1: 87

2: 89.4

3: 90.3

4: 92.1

5: 92.1

6: 95

7: 98.6

8: 100

9: 101.7

10: 104.8

So 5th value: 92.1, 6th: 95. So median_girls = (92.1 + 95)/2 = 93.55? Wait, no, wait, 92.1 + 95 = 187.1, divided by 2 is 93.55. Wait, but boys' median was 98.55. So the difference in medians: 98.55 - 93.55 = 5 cm.

Step3: Find the Interquartile Range (IQR) for boys and girls

IQR is Q3 - Q1, where Q1 is the median of the lower half, Q3 is the median of the upper half.

For boys (n=10), the lower half is the first 5 data points: 85.2, 86.7, 92.4, 93.8, 97.5. The median of the lower half (Q1) is the 3rd value (since n=5, odd), so Q1_boys = 92.4.

The upper half is the last 5 data points: 99.6, 101.3, 102.9, 105.1, 107. The median of the upper half (Q3) is the 3rd value, so Q3_boys = 102.9.

Thus, IQR_boys = Q3 - Q1 = 102.9 - 92.4 = 10.5 cm.

For girls (n=10), lower half: first 5 data points: 87, 89.4, 90.3, 92.1, 92.1. Median of lower half (Q1) is 3rd value: 90.3.

Upper half: last 5 data points: 95, 98.6, 100, 101.7, 104.8. Median of upper half (Q3) is 3rd value: 100.

Thus, IQR_girls = Q3 - Q1 = 100 - 90.3 = 9.7 cm (approximately 10 cm, or 9.7, close to 10).

Wait, but let's check again. Wait, for girls, lower half: positions 1-5: 87, 89.4, 90.3, 92.1, 92.1. The median (Q1) is the 3rd value: 90.3. Upper half: positions 6-10: 95, 98.6, 100, 101.7, 104.8. Median (Q3) is 3rd value: 100. So IQR_girls = 100 - 90.3 = 9.7 ≈ 10 cm.

For boys, lower half: positions 1-5: 85.2, 86.7, 92.4, 93.8, 97.5. Median (Q1) is 3rd value: 92.4. Upper half: positions 6-10: 99.6, 101.3, 102.9, 105.1, 107. Median (Q3) is 3rd value: 102.9. So IQR_boys = 102.9 - 92.4 = 10.5 ≈ 10 cm (or 10.5).

So both IQRs are approximately 10 cm (boys: ~10.5, girls: ~9.7, average around 10).

The difference in medians: boys' median (98.55) - girls' median (93.55) = 5 cm.

Now, check the options:

Option A: difference of medians (5) is about one-half the IQR (10). 5 is half of 10. So 5 ≈ (1/2)*10. That matches.

Option B: one-fourth of IQR (10) is 2.5, but difference is 5, so no.

Options C and D: say medians cannot be compared, but we just compared them, so C and D are wrong.

Wait, but let's recheck the median calculations. Wait, maybe I made a mistake in sorting.

Wait, boys' data: original data points (before sorting):

105.1, 101.3, 86.7, 93.8, 92.4, 85.2, 99.6, 97.5, 102.9, 107.

Let's sort them correctly:

85.2, 86.7, 92.4, 93.8, 97.5, 99.6, 101.3, 102.9, 105.1, 107. Yes, that's correct (ascending order). So n=10, so median is average of 5th and 6th: 5th is 97.5, 6th is 99.6. So (97.5 + 99.6)/2 = 98.55. Correct.

Girls' data: original data points:

104.8, 87, 95, 92.1, 100, 90.3, 98.6, 101.7, 89.4, 92.1.

Sort them:

87, 89.4, 90.3, 92.1, 92.1, 95, 98.6, 100, 101.7, 104.8. Correct. So median is average of 5th (92.1) and 6th (95): (92.1 + 95)/2 = 93.55. Correct.

Difference in medians: 98.55 - 93.55 = 5.

IQR for boys: Q1 is median of first 5 (positions 1-5: 85.2, 86.7, 92.4, 93.8, 97.5). Median of these 5 is the 3rd value: 92.4. Q3 is median of last 5 (positions 6-10: 99.6, 101.3, 102.9, 105.1, 107). Median is 3rd value: 102.9. So IQR_boys = 102.9 - 92.4 = 10.5.

IQR for girls: Q1 is median of first 5 (positions 1-5: 87, 89.4, 90.3, 92.1, 92.1). Median is 3rd value: 90.3. Q3 is median of last 5 (positions 6-10: 95, 98.6, 100, 101.7, 104.8). Median is 3rd value: 100. So IQR_girls = 100 - 90.3 = 9.7 ≈ 10.

So the IQR is approximately 10 (for boys ~10.5, girls ~9.7, so average ~10). The difference in medians is 5, which is half of 10. So the difference of the medians is about one-half the interquartile range of either data set. So option A is correct.

Answer:

A. The difference of the medians is about one-half the interquartile range of either data set.