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
[tex]\begin{gathered} \operatorname{mean}=\frac{\sum ^{}_{}fx}{\sum ^{}_{}f}=\frac{178}{20}=8.9 \\ \operatorname{mean}=8.9 \end{gathered}[/tex]Now, if the game with 3 hits is removed, the mean becomes
So, we will ignore the line with 3 hits we have
[tex]\begin{gathered} \operatorname{mean}=\frac{\sum^{}_{}fx}{\sum^{}_{}f}=\frac{175}{19}=9.2 \\ \operatorname{mean}=9.2 \end{gathered}[/tex]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
[tex]\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}[/tex]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.
