Question 2 (Multiple Choice, Worth 2 Points)

The table shows the number of goals made by two hockey players.

\begin{tabular}{|c|c|}
\hline Player A & Player B \\
\hline [tex]$2,3,1,3,2,2,1,3,6$[/tex] & [tex]$1,4,5,1,2,4,5,5,11$[/tex] \\
\hline
\end{tabular}

Find the best measure of variability for the data and determine which player was more consistent.

A. Player B is the most consistent, with an IQR of 3.5.
B. Player B is the most consistent, with a range of 10.
C. Player A is the most consistent, with an IQR of 1.5.
D. Player A is the most consistent, with a range of 5.



Answer :

Let's analyze the given data to determine the best measure of variability and find which player was more consistent.

Firstly, we need to calculate the range and the Interquartile Range (IQR) for both players.

### Range Calculation

The range is calculated as follows:
Range = Maximum value - Minimum value

For Player A:
- Maximum value = 6
- Minimum value = 1
Range for Player A = 6 - 1 = 5

For Player B:
- Maximum value = 11
- Minimum value = 1
Range for Player B = 11 - 1 = 10

### IQR Calculation

The IQR is calculated by first finding the first quartile (Q1) and the third quartile (Q3), and then subtracting Q1 from Q3.
IQR = Q3 - Q1

For Player A:
- First Quartile (Q1) is the 25th percentile.
- Third Quartile (Q3) is the 75th percentile.

Using the given data, these values can be found approximately. For Player A:
- Q1 (25th percentile) ≈ 2
- Q3 (75th percentile) ≈ 3
IQR for Player A = 3 - 2 = 1

For Player B:
- First Quartile (Q1) is the 25th percentile.
- Third Quartile (Q3) is the 75th percentile.

For Player B:
- Q1 (25th percentile) ≈ 2
- Q3 (75th percentile) ≈ 5
IQR for Player B = 5 - 2 = 3

### Determining Consistency

The player with a smaller IQR is more consistent because the IQR measures the spread of the middle 50% of the data.

For Player A:
- IQR = 1

For Player B:
- IQR = 3

### Conclusion

Player A has a smaller IQR. Therefore, Player A is more consistent.

Answer: Player A is the most consistent, with an IQR of 1.5.