Question

The leaders of two youth sports teams randomly select five members of each team and record their ages, as shown.

Answer 1

Answer:

Members of Team A are likely older, and they have less variability in their ages.

Step-by-step explanation:

To solve this question, we need to find the mean and the standard deviation for each team.

Team A 16, 13, 12, 16, 13

The mean is:

$M = \frac{16 + 13 + 12 + 16 + 13}{5} = 14$

The standard deviation is:

$S = \sqrt{\frac{(16-14)^2 + (13-14)^2 + (12-14)^2 + (16-14)^2 + (13-14)^2}{5}} = 1.67$

Team B: 10, 13, 16, 16, 10

The mean is:

$M = \frac{10 + 13 + 16 + 16 + 10}{5} = 13$

The standard deviation is:

$S = \sqrt{\frac{(10-13)^2 + (13-13)^2 + (16-13)^2 + (16-13)^2 + (10-13)^2}{5} = 13} = 2.68$

Which statement appropriately compares the ages of the team members?

Sample of team A has higher mean age, which means that they are likely older. They also have a smaller standard deviation, so they likely have less variability in their ages.