Respuesta :

xKelvin

Answer:

D

Step-by-step explanation:

Let's first find the mean and median of each class.

Class 1:

The mean is simply all the numbers added up and then divided by the number of elements. There are 9 students in Class 1. Thus, we add all the ages up and then divide by 9. Thus:

[tex]\text{Class 1 Mean }= \frac{14+15+15+16+16+16+17+17+18}{9} \\=144/9=16[/tex]

The median is simply the middle number when the data sets are placed in order. The median of Class 1 is 16, the number in the middle.

Class 2:

Again, Class 2 has 9 students. Add up all the ages and then divide:

[tex]\text{Class 2 Mean }= \frac{13+14+15+16+16+17+18+18+19}{9}\\ =146/9\approx16.2222[/tex]

The median is the middle number of the data set. The median of Class 2 is 16.

Therefore, the mean of Class 2 is larger than the mean of Class 1. The medians of the two classes are equivalent.

Of the answer choices given, only D is correct.

wegnerkolmp2741o

Answer:

The mean of class II is larger and the median is the same

Step-by-step explanation:

Class I

14,15,15,16,16,16,17,17,18

The mean is

(14+15+15+16+16+16+17+17+18)/9

144/9 = 16

The median is the middle number

14,15,15,16,    16,    16,17,17,18

median = 16

Class II

13,14,15,16,16,17,18,18,19

The mean is

(13+14+15+16+16+17+18+18+19)/9

146/9 = 16.2repeating

The median is the middle number

13,14,15,16    ,16,       17,18,18,19

median = 16

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