paper planes\n1. collect data from your groups paper airplanes:\ngroup a\n| name | distance (ft) |\n| ---- |…
paper planes\n1. collect data from your groups paper airplanes:\ngroup a\n| name | distance (ft) |\n| ---- | ---- |\n| walker | 15.8 ft |\n| bryson | 25.1 ft |\n| jamon | 25.8 ft |\n| beyonca | 15.9 ft |\ngroup b\n| name | distance (ft) |\n| ---- | ---- |\n| lily | 20.9 ft |\n| jalasa | 24.2 ft |\n| jettner | 19.2 ft |\n| jakari | 16.9 ft |\n2. find the following information for each group:\ngroup a\n| mean | mad | standard deviation | iqr |\n| ---- | ---- | ---- | ---- |\n| | | | |\n| min | q1 | median | q3 | max |\n| ---- | ---- | ---- | ---- | ---- |\n| | | | | |\ngroup b\n| mean | mad | standard deviation | iqr |\n| ---- | ---- | ---- | ---- |\n| | | | |\n| min | q1 | median | q3 | iqr |\n| ---- | ---- | ---- | ---- | ---- |\n| | | | | |\n3. draw a box plot for each group.\ngroup a\ngroup b\n4. what is the shape of distribution of your data?(your group)\n5. interpret the mean in relation to the problem.(your group)
Answer
Explanation:
Step1: Calculate mean for Group A
Group A data: 15.8, 25.1, 25.8, 15.9. Mean $\bar{x}_A=\frac{15.8 + 25.1+25.8+15.9}{4}=\frac{82.6}{4}=20.65$
Step2: Calculate mean - absolute - deviation (MAD) for Group A
Find absolute - deviations from the mean: $|15.8 - 20.65| = 4.85$, $|25.1-20.65| = 4.45$, $|25.8 - 20.65| = 5.15$, $|15.9 - 20.65| = 4.75$. MAD for Group A: $\frac{4.85 + 4.45+5.15+4.75}{4}=\frac{19.2}{4}=4.8$
Step3: Calculate standard deviation for Group A
First, find squared - deviations: $(15.8 - 20.65)^2=23.5225$, $(25.1 - 20.65)^2 = 19.8025$, $(25.8 - 20.65)^2=26.5225$, $(15.9 - 20.65)^2 = 22.5625$. Variance $s_A^2=\frac{23.5225 + 19.8025+26.5225+22.5625}{4 - 1}=\frac{92.41}{3}\approx30.8033$. Standard deviation $s_A=\sqrt{30.8033}\approx5.55$
Step4: Calculate quartiles for Group A
Arrange data in ascending order: 15.8, 15.9, 25.1, 25.8. Median (Q2) is $\frac{15.9 + 25.1}{2}=20.5$. Q1 is 15.85, Q3 is 25.45, Min is 15.8, Max is 25.8, IQR for Group A is $25.45−15.85 = 9.6$
Step5: Calculate mean for Group B
Group B data: 20.9, 24.2, 19.2, 16.9. Mean $\bar{x}_B=\frac{20.9+24.2 + 19.2+16.9}{4}=\frac{81.2}{4}=20.3$
Step6: Calculate MAD for Group B
Find absolute - deviations from the mean: $|20.9 - 20.3| = 0.6$, $|24.2-20.3| = 3.9$, $|19.2 - 20.3| = 1.1$, $|16.9 - 20.3| = 3.4$. MAD for Group B: $\frac{0.6+3.9 + 1.1+3.4}{4}=\frac{9}{4}=2.25$
Step7: Calculate standard deviation for Group B
First, find squared - deviations: $(20.9 - 20.3)^2 = 0.36$, $(24.2 - 20.3)^2=15.21$, $(19.2 - 20.3)^2 = 1.21$, $(16.9 - 20.3)^2 = 11.56$. Variance $s_B^2=\frac{0.36+15.21+1.21+11.56}{4 - 1}=\frac{28.34}{3}\approx9.4467$. Standard deviation $s_B=\sqrt{9.4467}\approx3.07$
Step8: Calculate quartiles for Group B
Arrange data in ascending order: 16.9, 19.2, 20.9, 24.2. Median (Q2) is $\frac{19.2 + 20.9}{2}=20.05$. Q1 is 18.05, Q3 is 22.55, Min is 16.9, Max is 24.2, IQR for Group B is $22.55−18.05 = 4.5$
Group A Results
| Mean | MAD | Standard Deviation | IQR |
|---|---|---|---|
| 20.65 | 4.8 | 5.55 | 9.6 |
| Min | Q1 | Median | Q3 |
| 15.8 | 15.85 | 20.5 | 25.45 |
Group B Results
| Mean | MAD | Standard Deviation | IQR |
|---|---|---|---|
| 20.3 | 2.25 | 3.07 | 4.5 |
| Min | Q1 | Median | Q3 |
| 16.9 | 18.05 | 20.05 | 22.55 |
- To draw box - plots:
- For Group A: Mark Min at 15.8, Q1 at 15.85, Median at 20.5, Q3 at 25.45, and Max at 25.8 on a number line. Draw a box from Q1 to Q3 with a line at the median. Draw whiskers from the box to Min and Max.
- For Group B: Mark Min at 16.9, Q1 at 18.05, Median at 20.05, Q3 at 22.55, and Max at 24.2 on a number line. Draw a box from Q1 to Q3 with a line at the median. Draw whiskers from the box to Min and Max.
- Shape of distribution for Group A: Since the mean (20.65) is greater than the median (20.5) and the right - tail (from median to Max) is longer than the left - tail (from Min to median), the distribution of Group A is right - skewed.
- Shape of distribution for Group B: Since the mean (20.3) is close to the median (20.05) and the lengths of the tails are relatively similar, the distribution of Group B is approximately symmetric.
- Interpretation of mean for Group A: The mean distance of 20.65 feet represents the average distance that the paper airplanes in Group A flew.
- Interpretation of mean for Group B: The mean distance of 20.3 feet represents the average distance that the paper airplanes in Group B flew.
Answer:
Group A results are shown in the Group A tables above, and Group B results are shown in the Group B tables above. Box - plots can be drawn as described. Group A has a right - skewed distribution, Group B has an approximately symmetric distribution. The mean for Group A represents the average flight distance of their paper airplanes as 20.65 feet, and for Group B as 20.3 feet.