Which statement correctly compares the centers of the distributions?


A. The mean of Southview HS is greater than the mean of North
Heights HS.

B. The range of North Heights HS is greater than the range of
Southview HS.

C. The median of Southview HS is greater than the median of North
Heights HS.

D. The mean of North Heights HS is greater than the mean of
Southview HS.

Which statement correctly compares the centers of the distributions A The mean of Southview HS is greater than the mean of North Heights HS B The range of North class=

Respuesta :

Answer:  B

Step-by-step explanation:

mean - the average of a set of numbers.

To find the mean of a data set, add up all of the numbers in the set, and then divide that total by the number of numbers in the set.

range - The difference between the lowest and highest values.

Take the highest value and subtract the lowest value to find range.

median - the middle number in a set, also the number that is halfway into the set.

Arrange all numbers (even if the same number list it twice)  in order from smallest to largest.  mark out one at each end working way towards middle.

I started by finding range (B) because this took less time, it was a true statement.

The statement that best compares the centers of distributions of North Heights HS and Southview HS is: D. The mean (33) of North Heights HS is greater than the mean (29) of Southview HS.

Recall:

  • Center of a distribution describes the middle of a data set.
  • Mean and Median are two measures that can give us an idea about the center of a distribution.

Thus, let's estimate the mean and median of the distribution shown in the histogram to determine the correct statement that describes the centers of both data distribution.

Mean of North Heights HS:

An estimate of the mean = sum of the middle value of each class size x frequency of the class / the total number of data value

Sum of the middle value of each class size x frequency of the class = [tex]29 \times 4 + 31 \times 8 + 33 \times 5 + 35 \times 4 + 37 \times 2 + 39 \times 1 + 43 \times 1 = 825[/tex]

The total number of data value = 4 + 8 + 5 + 4 + 2 + 1 + 1 = 25

Mean = [tex]\frac{825}{25} = \mathbf{33}[/tex]

Mean of Southview HS:

Sum of the middle value of each class size x frequency of the class = [tex]25 \times 2 + 27 \times 6 + 29 \times 9 + 31 \times 6 + 33 \times 2 = 725[/tex]

The total number of data value = 2 + 6 + 9 + 6 + 2 = 25

Mean = [tex]\frac{725}{25} = \mathbf{29}[/tex]

Comparing the means: the mean of North Heights HS , which is 33, is greater than the mean of Southview, which is 29.

Median of North Heights HS:

The median is the 13th data value which lies between the class interval of 32 - 34.

The median will fall within that range of values.

Median of Southview HS:

The median is the 13th data value which lies between the class interval of 28 - 30.

Therefore, we can safely say the median value for North Heights HS is greater than the median value for Southview HS.

In conclusion, the statement that best compares the centers of distributions of North Heights HS and Southview HS is: D. The mean (33) of North Heights HS is greater than the mean (29) of Southview HS.

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