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Sagot :
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:
[tex]M = \frac{16 + 13 + 12 + 16 + 13}{5} = 14[/tex]
The standard deviation is:
[tex]S = \sqrt{\frac{(16-14)^2 + (13-14)^2 + (12-14)^2 + (16-14)^2 + (13-14)^2}{5}} = 1.67[/tex]
Team B: 10, 13, 16, 16, 10
The mean is:
[tex]M = \frac{10 + 13 + 16 + 16 + 10}{5} = 13[/tex]
The standard deviation is:
[tex]S = \sqrt{\frac{(10-13)^2 + (13-13)^2 + (16-13)^2 + (16-13)^2 + (10-13)^2}{5} = 13} = 2.68[/tex]
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.
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