Question

paper planes

1. collect data from your groups paper airplanes:

group a

name	distance (ft)
bryson	25.1 ft
jamon	25.8 ft
beyonca	15.9 ft

group b

name	distance (ft)
jalasa	24.2 ft
jettner	19.2 ft
jakari	16.9 ft

1. find the following information for each group:

group a

mean	mad	standard deviation	iqr
min	q1	median	q3	max

group b

mean	mad	standard deviation	iqr
min	q1	median	q3	iqr

1. draw a box plot for each group.

group a
group b

1. what is the shape of distribution of your data?(your group)
2. interpret the mean in relation to the problem.(your group)

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
Min	Q1	Median	Q3	Max
15.8	15.85	20.5	25.45	25.8

Group B Results

Mean	MAD	Standard Deviation	IQR
Min	Q1	Median	Q3	Max
16.9	18.05	20.05	22.55	24.2

1. 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.

1. 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.

1. 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.