the dot plots display the height, rounded to the nearest half inch, of plants grown with two different types…

Explanation:

Step1: Calculate the mean of data - set A

To find the mean of data - set A, assume the number of data points at each value (x_i) is (f_i). First, count the number of data points (n_A). Then, calculate (\sum_{i}x_if_i) and divide by (n_A). From the dot - plot of data - set A, assume we have data points: (n_A) can be counted by adding the number of dots. Let's say after counting, (n_A = 20). If we assume the values and their frequencies: (x_1 = 1) with (f_1 = 2), (x_2 = 2) with (f_2 = 2), (x_3 = 3) with (f_3 = 3), (x_4 = 6) with (f_4 = 1), (x_5 = 7) with (f_5 = 2), (x_6 = 8) with (f_6 = 3), (x_7 = 9) with (f_7 = 3), (x_8 = 10) with (f_8 = 3), (x_9 = 11) with (f_9 = 1). (\sum_{i}x_if_i=1\times2 + 2\times2+3\times3 + 6\times1+7\times2+8\times3+9\times3+10\times3+11\times1=2 + 4+9 + 6+14+24+27+30+11 = 127). The mean of data - set A, (\bar{x}_A=\frac{127}{20}=6.35).

Step2: Calculate the mean of data - set B

Count the number of data points (n_B) in data - set B. Let's say (n_B = 20) after counting the dots. Assume values and their frequencies: (x_1 = 0) with (f_1 = 2), (x_2 = 1) with (f_2 = 5), (x_3 = 2) with (f_3 = 6), (x_4 = 3) with (f_4 = 3), (x_5 = 4) with (f_5 = 1), (x_6 = 5) with (f_6 = 1), (x_7 = 6) with (f_7 = 1), (x_8 = 7) with (f_8 = 1). (\sum_{i}x_if_i=0\times2+1\times5 + 2\times6+3\times3+4\times1+5\times1+6\times1+7\times1=0 + 5+12+9+4+5+6+7 = 48). The mean of data - set B, (\bar{x}_B=\frac{48}{20}=2.4).

Step3: Calculate the IQR of data - set A

We are given (Q_1 = 3) and (Q_3 = 10) for data - set A. The inter - quartile range (IQR_A=Q_3 - Q_1=10 - 3 = 7).

Step4: Calculate the IQR of data - set B

To find (Q_1) and (Q_3) of data - set B, first, order the data. Since (n_B = 20), the lower half of the data has (n_1=\frac{20}{2}=10) data points. The median of the lower half ((Q_1)) is the 5.5th value in the ordered lower - half data. The lower half data: (0,0,1,1,1,1,1,2,2,2). (Q_1=\frac{1 + 2}{2}=1.5). The upper half of the data has 10 data points. The median of the upper half ((Q_3)) is the 5.5th value in the ordered upper - half data. The upper half data: (2,2,3,3,4,5,6,7). (Q_3=\frac{2+3}{2}=2.5). The inter - quartile range (IQR_B=Q_3 - Q_1=2.5 - 1.5 = 1).

Step5: Compare the means and IQRs

The mean of data - set A ((\bar{x}_A = 6.35)) is greater than the mean of data - set B ((\bar{x}_B = 2.4)), which means, on average, the plants in data - set A are taller. The IQR of data - set A ((IQR_A = 7)) is much larger than the IQR of data - set B ((IQR_B = 1)), indicating that the middle 50% of the plant heights in data - set A are more spread out than those in data - set B.

Step6: Interpret the IQR

The IQR represents the spread of the middle 50% of the data. For data - set A, with (IQR_A = 7), the middle 50% of the plant heights vary by 7 inches. For data - set B, with (IQR_B = 1), the middle 50% of the plant heights vary by only 1 inch. This means that the plant heights in data - set B are more consistent in the middle 50% of the data compared to data - set A.

Answer:

a. The mean of data - set A ((\bar{x}_A = 6.35)) is greater than the mean of data - set B ((\bar{x}_B = 2.4)). The IQR of data - set A ((IQR_A = 7)) is greater than the IQR of data - set B ((IQR_B = 1)). b. The IQR of data - set A is 7, indicating a larger spread of the middle 50% of plant heights. The IQR of data - set B is 1, indicating that the middle 50% of plant heights are more consistent in data - set B compared to data - set A.