Question

the dot plots display the height, rounded to the nearest half inch, of plants grown with two different types of fertilizer.
a)
0 1 2 3 4 5 6 7 8 9 10 11 12
height (inches)
q1 = 3
q3 = 10
b)
0 1 2 3 4 5 6 7 8 9 10 11 12
height (inches)
a. compare the mean and iqr of the two data sets.
b. what does the iqr tell you about the two groups of plants?

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.