the mean ages with standard deviations of four swim teams at a swim club are given below. team mean standard…

the mean ages with standard deviations of four swim teams at a swim club are given below. team mean standard deviation stars 16 4.1 dolphins 18 1.5 giants 14 0.3 mackerels 15 2.3 which statement is most likely to be true? the ages of the mackerels are the most dispersed from the teams mean. the ages of the stars are the most dispersed from the teams mean. the ages of the dolphins are the most dispersed from the teams mean. the ages of the giants are the most dispersed from the teams mean.

the mean ages with standard deviations of four swim teams at a swim club are given below. team mean standard deviation stars 16 4.1 dolphins 18 1.5 giants 14 0.3 mackerels 15 2.3 which statement is most likely to be true? the ages of the mackerels are the most dispersed from the teams mean. the ages of the stars are the most dispersed from the teams mean. the ages of the dolphins are the most dispersed from the teams mean. the ages of the giants are the most dispersed from the teams mean.

Answer

Explanation:

Step1: Recall standard - deviation concept

Standard deviation measures data dispersion from the mean. Larger value means more dispersion.

Step2: Compare standard - deviation values

We have: Stars ($\sigma_{Stars}=4.1$), Dolphins ($\sigma_{Dolphins}=1.5$), Giants ($\sigma_{Giants}=0.3$), Mackerels ($\sigma_{Mackerels}=2.3$). Since $4.1>2.3 > 1.5>0.3$, Stars have the largest standard deviation.

Answer:

The ages of the Stars are the most dispersed from the team's mean.