Question

the table shows the daily time, in minutes, maggie spent at swim practice and band practice over the last ten days.
practice times
swim 55 50 70 75 60 55 55 75 65 70
band 45 60 90 35 40 50 90 40 45 50
which statement is true about the measures of center of the practice times?
○ maggie’s daily time at band practice tends to be greater than her time at swim practice.
○ maggie’s daily time at band practice has less variation than her daily time at swim practice.
○ maggie’s daily time at swim practice tends to be greater than her time at band practice.
○ maggie’s daily time at swim practice tends to have more variation than her daily time at band practice.

Explanation:

To solve this, we analyze the measures of center (mean, median) and variation (range, interquartile range) for swim and band practice times.

Step 1: Organize Swim Practice Data

Swim times: \( 50, 55, 55, 55, 60, 65, 70, 70, 75, 75 \) (sorted)

• Median: Middle value of 10 data points. Average of 5th and 6th: \( \frac{60 + 65}{2} = 62.5 \)
• Mean: \( \frac{50 + 55 + 55 + 55 + 60 + 65 + 70 + 70 + 75 + 75}{10} = \frac{630}{10} = 63 \)
• Range: \( 75 - 50 = 25 \)

Step 2: Organize Band Practice Data

Band times: \( 35, 40, 40, 45, 45, 50, 50, 60, 90, 90 \) (sorted)

• Median: Average of 5th and 6th: \( \frac{45 + 50}{2} = 47.5 \)
• Mean: \( \frac{35 + 40 + 40 + 45 + 45 + 50 + 50 + 60 + 90 + 90}{10} = \frac{545}{10} = 54.5 \)
• Range: \( 90 - 35 = 55 \)

Step 3: Compare Measures

• Center (Median/Mean): Swim’s median (62.5) and mean (63) are greater than Band’s (47.5, 54.5). So swim practice time tends to be greater.
• Variation (Range): Band’s range (55) is greater than Swim’s (25), so Band has more variation.

• Analyzed median, mean, and range for both datasets.
• Swim’s center (median/mean) is higher than Band’s.
• Band’s range (variation) is larger than Swim’s.

Answer:

Maggie’s daily time at swim practice tends to be greater than her time at band practice.