Question

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 team’s mean. The ages of the Stars are the most dispersed from the team’s mean. The ages of the Dolphins are the most dispersed from the team’s mean. The ages of the Giants are the most dispersed from the team’s mean.

Answer 1

Using the coefficient of variation, it is found that the correct statement is:

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

What is the coefficient of variation of a data-set?

The coefficient of variation of a data-set is given by the division of the standard deviation of the data-set by the mean of the data-set. The higher the coefficient, the most dispersed the data-set is from the mean.

For the Stars, the mean is of 16 and the standard deviation is of 4.1, hence the coefficient is:

$cv_S = \frac{4.1}{16} = 0.256$

For the Dolphins, the mean is of 18 and the standard deviation is of 1.5, hence the coefficient is:

$cv_D = \frac{1.5}{18} = 0.083$

For the Giants, the mean is of 14 and the standard deviation is of 0.3, hence the coefficient is:

$cv_G = \frac{0.3}{14} = 0.021$

For the Mackerels, the mean is of 15 and the standard deviation is of 2.3, hence the coefficient is:

$cv_M = \frac{2.3}{15} = 0.153$

Due to the higher coefficient, the ages of the Stars are the most dispersed from the team's mean.

You can learn more about the coefficient of variation at https://brainly.com/question/24495359

Answer 2

Answer:

Statement 2 (The ages of the Stars are the most dispersed from the team’s mean).

Step-by-step explanation:

Edge