Question

Lucy and Charley are bowling. Charley bowled already and had a mean absolute deviation of 7. Lucy bowls and gets the following scores: 115, 110, 123, 118, 120. Who is a more consistent bowler?O LucyO Charley

Answer 1

We have to find who is a more consistent bowler.

This corresponds to the one that has the least spread in their scores.

As we already know the mean absolute deviation for Charley's scores, we will use the same measure of spread for the scores of Lucy.

We then first have to calculate the mean scores for Lucy:

`\begin{gathered} M=(1)/(5)(115+110+123+118+120) \\ M=(1)/(5)(586) \\ M=117.2 \end{gathered}`

Now, we can calculate the mean absolute deviation as:

`\begin{gathered} MAD=(1)/(5)\sum^|x_i-M| \\ MAD=(1)/(5)(|115-117.2|+|110-117.2|+|123-117.2|+|118-117.2|+|120-117.2|) \\ MAD=(1)/(5)(2.2+7.2+5.8+0.8+2.8) \\ MAD=(1)/(5)(18.8) \\ MAD=3.76 \end{gathered}`

As Lucy's scores have a smaller MAD than Charley's scores, we can consider that Lucy is a more consistent bowler.

Answer: Lucy