Question

5. The dot plot displays the number of hits a baseball team made in severalgames.The distribution is skewed to the left.OOOO....wt324.13125un0010116 7 8 9hits per game5a. If the game with 3 hits is considered to be recorded in error, it might be pointremoved from the data set. If that happens, what happens to the mean ofthe data set?

Answer 1

SOLUTION

First let us find the mean of the data set with the game of 3 hits included.

We will make a frequency table to do this, we have

From the table, mean is calcuated as

$\begin{gathered} \operatorname{mean}=\frac{\sum ^{}_{}fx}{\sum ^{}_{}f}=\frac{178}{20}=8.9 \\ \operatorname{mean}=8.9 \end{gathered}$

Now, if the game with 3 hits is removed, the mean becomes

So, we will ignore the line with 3 hits we have

$\begin{gathered} \operatorname{mean}=\frac{\sum^{}_{}fx}{\sum^{}_{}f}=\frac{175}{19}=9.2 \\ \operatorname{mean}=9.2 \end{gathered}$

Hence if the game with 3 hits is removed, the mean increases from 8.9 to 9.2

Now, for the median

From the table, let's trace the median,

From the picture above, if we consider 3 hits, looking at the frequencies, we will add the first parts of the frequencies, then the last parts.

The first

$\begin{gathered} 1+2+4=7 \\ \text{last part } \\ 4+4=8 \\ so\text{ median should be in the frequency of 5, which is ideally the middle} \\ \text{trancing the number we get 9} \end{gathered}$

So, with 3 hits included, the median is 9.

With 3 hits removed

The upper part of the frequency is 2 + 4 = 6

The lower part 4 + 4 = 8

So the median will still fall in the frequency part of 5, making the median 9.

Hence if the game with 3 hits is removed, the median remains the same, that is 9.