Question

the table shows the weights of giant pumpkins grown in two towns, cedartown and baytown. match each phrase to its value.

weight of cedartown pumpkins (pounds)	weight of baytown pumpkins (pounds)
753	622
575	713.5
685.5	579
718	644

20.2 40 40.2 60.4 575 628.5 668.5 713.5

• the average weight of pumpkins grown in cedartown
• the average weight of pumpkins grown in baytown
• the mean absolute deviation of the pumpkin weights in cedartown
• the mean absolute deviation of the pumpkin weights in baytown

Explanation:

Step1: Calculate Cedar - town average

Sum of Cedar - town weights: $611 + 753+575 + 685.5+718=3342.5$. Number of data points $n = 5$. Average $\bar{x}_{Cedar - town}=\frac{3342.5}{5}=668.5$.

Step2: Calculate Bay - town average

Sum of Bay - town weights: $584+622 + 713.5+579+644 = 3142.5$. Number of data points $n = 5$. Average $\bar{x}_{Bay - town}=\frac{3142.5}{5}=628.5$.

Step3: Calculate mean - absolute - deviation for Cedar - town

Deviations from the mean for Cedar - town: $|611 - 668.5|=57.5$, $|753 - 668.5| = 84.5$, $|575 - 668.5|=93.5$, $|685.5 - 668.5| = 17$, $|718 - 668.5|=49.5$. Sum of absolute deviations $=57.5 + 84.5+93.5+17+49.5 = 302$. Mean - absolute - deviation $MAD_{Cedar - town}=\frac{302}{5}=60.4$.

Step4: Calculate mean - absolute - deviation for Bay - town

Deviations from the mean for Bay - town: $|584 - 628.5|=44.5$, $|622 - 628.5| = 6.5$, $|713.5 - 628.5|=85$, $|579 - 628.5|=49.5$, $|644 - 628.5|=15.5$. Sum of absolute deviations $=44.5+6.5 + 85+49.5+15.5 = 201.5$. Mean - absolute - deviation $MAD_{Bay - town}=\frac{201.5}{5}=40.3\approx40$.

Answer:

the average weight of pumpkins grown in Cedartown: 668.5
the average weight of pumpkins grown in Baytown: 628.5
the mean absolute deviation of the pumpkin weights in Cedartown: 60.4
the mean absolute deviation of the pumpkin weights in Baytown: 40