MATH SOLVE

7 months ago

Q:
# Enzo and Zahra are playing a game. Their scores for five games are shown in the table below. Enzo’s and Zahra’s Scores Enzo’s Scores 39 47 46 50 38 Zahra’s Scores 37 33 45 30 45 Enzo’s mean score is 44. Zahra’s mean score is 38. The difference between the mean scores is about how many times the mean absolute deviation of the data sets, to the nearest whole number?

Accepted Solution

A:

Given the table that contains Enzo's and Zahra's scores for five games, we solve the required problem as follows:

[tex]\begin{tabular} {|c|c|c|c|} Enzo's scores&Zahra's scores&$|x-\bar{x}|$&$|y-\bar{y}|$\\ (x)&(y)&&\\[1ex]39&37&5&1\\ 47&33&3&5\\ 46&45&2&7\\ 50&30&6&8\\ 38&45&6&7\\ $\bar{x}= \frac{220}{5} =44$&$\bar{x}=\frac{190}{5} =38$&MAD$=\frac{22}{5}=4.4$&MAD=$\frac{28}{5}=5.6 \end{tabular}[/tex]

The difference between the mean scores is 44 - 38 = 6

The difference between the mean scores is about the same size as the mean absolute deviation of the data sets.

[tex]\begin{tabular} {|c|c|c|c|} Enzo's scores&Zahra's scores&$|x-\bar{x}|$&$|y-\bar{y}|$\\ (x)&(y)&&\\[1ex]39&37&5&1\\ 47&33&3&5\\ 46&45&2&7\\ 50&30&6&8\\ 38&45&6&7\\ $\bar{x}= \frac{220}{5} =44$&$\bar{x}=\frac{190}{5} =38$&MAD$=\frac{22}{5}=4.4$&MAD=$\frac{28}{5}=5.6 \end{tabular}[/tex]

The difference between the mean scores is 44 - 38 = 6

The difference between the mean scores is about the same size as the mean absolute deviation of the data sets.