Question

select all the correct answers. the table records the rainfall in two cities during a particular week. which statements about the data are true? rainfall in mckesson (inches) rainfall in sansome (inches) 3.76 3.14 2.45 3.01 3.01 2.30 3.82 2.26 2.32 2.12 3.97 2.71 1.74 1.97 their means cannot be compared based on their mean absolute deviations because the mean absolute deviation for mckessons rainfall is nearly twice that of sansomes. their means cannot be compared based on their mean absolute deviations because the mean absolute deviation for sansomes rainfall is nearly twice that of mckessons. the difference of the means of the two data is 0.5. the difference of the means of the two cities is 1.

Explanation:

Step1: Calculate McKesson mean

\(\bar{x}_{McKesson}=\frac{3.76 + 2.45+3.01 + 3.82+2.32+3.97+1.74}{7}=3.01\)

Step2: Calculate Sansome mean

\(\bar{x}_{Sansome}=\frac{3.14 + 3.01+2.30+2.26+2.12+2.71+1.97}{7}=2.50\)

Step3: Calculate mean difference

\(3.01−2.50 = 0.5\)

Step4: Calculate McKesson MAD

\(MAD_{McKesson}=\frac{|3.76 - 3.01|+|2.45 - 3.01|+\cdots+|1.74 - 3.01|}{7}=0.72\)

Step5: Calculate Sansome MAD

\(MAD_{Sansome}=\frac{|3.14 - 2.50|+|3.01 - 2.50|+\cdots+|1.97 - 2.50|}{7}\approx0.39\)

Answer:

1. First, calculate the mean of rainfall in McKesson:

• The data for McKesson is \(3.76,2.45,3.01,3.82,2.32,3.97,1.74\).
• The sum of these values is \(3.76 + 2.45+3.01 + 3.82+2.32+3.97+1.74=21.07\).
• There are \(n = 7\) data - points. The mean \(\bar{x}_{McKesson}=\frac{21.07}{7}=3.01\).

1. Then, calculate the mean of rainfall in Sansome:

• The data for Sansome is \(3.14,3.01,2.30,2.26,2.12,2.71,1.97\).
• The sum of these values is \(3.14 + 3.01+2.30+2.26+2.12+2.71+1.97 = 17.51\).
• There are \(n = 7\) data - points. The mean \(\bar{x}_{Sansome}=\frac{17.51}{7}=2.5014\approx2.50\).

1. Calculate the difference of the means:

• The difference of the means is \(3.01−2.50 = 0.51\approx0.5\).

1. Now, calculate the mean - absolute - deviation (MAD):

• For McKesson:
• Calculate the absolute - deviations from the mean:
• \(|3.76 - 3.01|=0.75\), \(|2.45 - 3.01| = 0.56\), \(|3.01 - 3.01|=0\), \(|3.82 - 3.01| = 0.81\), \(|2.32 - 3.01|=0.69\), \(|3.97 - 3.01| = 0.96\), \(|1.74 - 3.01|=1.27\).
• The sum of the absolute - deviations is \(0.75+0.56 + 0+0.81+0.69+0.96+1.27 = 5.04\).
• The MAD for McKesson is \(\frac{5.04}{7}=0.72\).
• For Sansome:
• Calculate the absolute - deviations from the mean:
• \(|3.14 - 2.50| = 0.64\), \(|3.01 - 2.50|=0.51\), \(|2.30 - 2.50| = 0.20\), \(|2.26 - 2.50|=0.24\), \(|2.12 - 2.50| = 0.38\), \(|2.71 - 2.50|=0.21\), \(|1.97 - 2.50| = 0.53\).
• The sum of the absolute - deviations is \(0.64+0.51+0.20+0.24+0.38+0.21+0.53 = 2.71\).
• The MAD for Sansome is \(\frac{2.71}{7}\approx0.39\).
• The MAD of McKesson is not nearly twice that of Sansome, and vice - versa.

The correct statement is: The difference of the means of the two cities is \(0.5\).