Here is the height, in inches, of 10 randomly selected members of the girls' dance team and 10 randomly selected of the girls' volleyball team
Based on the interquartile ranges of the two sets of data, which is a reasonable conclusion concerning heights of the players on the two teams?
Volley ball team:67,63,70,67,68,69.70,68,72,68
Dance team:62,67,59,63,67,60,66,60,66,63
A. there is a greater variability of heights for the middle 50% of the dance team than for the middle50% of the volleyball team members
B. There is a greater variability of heights for the middle 50% of the volleyball team than for the middle50% of the dance team members
c. The variability for the 50% was the same for both team members
D. the average height, in inches, was higher for the members of the dance than for the members of the volleyball team.
I had B. since 68 is the middle and the is 5 numbers before it to get to the lowest number,63, and 4 to get to the highest number,72.
In dance 63, 59 is the lowest nuber, being 4 away and the highest is 67, being 4 away., so volleyball team has more variability.
Is this correct. I don't know if I did this correctly, can you help me; thanks
You are on the right track, but your reasoning is a bit off. To determine the interquartile range (IQR) for each set of data, we need to find the values that divide the data into quartiles.
Let's start by finding the quartiles for each team:
Volleyball team:
- Arrange the data in ascending order: 63, 67, 68, 68, 69, 70, 70, 72
- The median (Q2) is the middle value, which is 68.
- To find Q1, we take the median of the lower half of the data: 63, 67, 68. The median of this set is 67.
- To find Q3, we take the median of the upper half of the data: 68, 69, 70, 70, 72. The median of this set is 70.
Dance team:
- Arrange the data in ascending order: 59, 60, 60, 62, 63, 63, 66, 66, 67, 67
- The median (Q2) is the middle value, which is 63.
- To find Q1, we take the median of the lower half of the data: 59, 60, 60, 62, 63. The median of this set is 60.
- To find Q3, we take the median of the upper half of the data: 63, 66, 66, 67, 67. The median of this set is 66.
Now we can calculate the IQR for each team by subtracting Q1 from Q3:
Volleyball team: IQR = Q3 - Q1 = 70 - 67 = 3 inches
Dance team: IQR = Q3 - Q1 = 66 - 60 = 6 inches
From the calculations, we can see that the IQR for the dance team (6 inches) is greater than the IQR for the volleyball team (3 inches). Therefore, the reasonable conclusion would be:
A. There is a greater variability of heights for the middle 50% of the dance team than for the middle 50% of the volleyball team members.
To determine the answer to this question, we need to calculate the interquartile range (IQR) for both the volleyball team and the dance team.
Steps to calculate the IQR:
1. Arrange the data in ascending order.
Volleyball team heights: 63, 67, 67, 68, 68, 69, 70, 70, 72
Dance team heights: 59, 60, 60, 62, 63, 63, 66, 66, 67, 67
2. Find the first quartile (Q1) and the third quartile (Q3). Q1 represents the 25th percentile, and Q3 represents the 75th percentile.
Volleyball team heights:
Q1 = (63 + 67) / 2 = 65 inches
Q3 = (69 + 70) / 2 = 69.5 inches
Dance team heights:
Q1 = (60 + 62) / 2 = 61 inches
Q3 = (66 + 67) / 2 = 66.5 inches
3. Calculate the IQR by subtracting Q1 from Q3.
Volleyball team IQR = 69.5 - 65 = 4.5 inches
Dance team IQR = 66.5 - 61 = 5.5 inches
Now, comparing the IQRs, we can make the conclusion.
In this case, the correct answer is A. There is a greater variability of heights for the middle 50% of the dance team than for the middle 50% of the volleyball team members. The dance team's IQR (5.5 inches) is greater than the volleyball team's IQR (4.5 inches), indicating more variability in heights on the dance team.
So, your initial intuition was correct. The volleyball team has less variability in heights for the middle 50% of its members than the dance team.