For a project, Fay kept track of the total annual amount of precipitation in inches for two cities over the last 12
years. The results are shown in the table below.
City A 41.50 69.43 48.15 39.23 37.03 66.02 47.62 68.38 38.69 52.48 70.03 43.74
City M 48.14 61.59 44.05 33.14 32.41 72.92 48.49 50.21 33.70 48.69 49.06 44.47
Find the mean and standard deviation for each city, and then compare the means and standard deviations. Round the means and standard deviations to two decimal places.
First, let's find the mean and standard deviation for each city:
For City A:
Mean = (41.50 + 69.43 + 48.15 + 39.23 + 37.03 + 66.02 + 47.62 + 68.38 + 38.69 + 52.48 + 70.03 + 43.74) / 12
Mean = 575.30 / 12
Mean = 47.94 inches
Standard deviation = sqrt[ Σ(xi - x̄)^2 / (n - 1) ]
Standard deviation = sqrt[ ( (41.50-47.94)^2 + (69.43-47.94)^2 + ... + (43.74-47.94)^2 ) / 11 ]
Standard deviation = sqrt[ (42.1956 + 441.0164 + ... + 6.9696) / 11 ]
Standard deviation = sqrt[ 703.4696 / 11 ]
Standard deviation = sqrt[63.9517818]
Standard deviation ≈ 7.99 inches
For City M:
Mean = (48.14 + 61.59 + 44.05 + 33.14 + 32.41 + 72.92 + 48.49 + 50.21 + 33.70 + 48.69 + 49.06 + 44.47) / 12
Mean = 526.67 / 12
Mean = 43.89 inches
Standard deviation = sqrt[ Σ(xi - x̄)^2 / (n - 1) ]
Standard deviation = sqrt[ ( (48.14-43.89)^2 + (61.59-43.89)^2 + ... + (44.47-43.89)^2 ) / 11 ]
Standard deviation = sqrt[ (17.5841 + 313.7521 + ... + 0.3364) / 11 ]
Standard deviation = sqrt[ 344.8036 / 11 ]
Standard deviation = sqrt[31.3457455]
Standard deviation ≈ 5.60 inches
Now, let's compare the means and standard deviations:
- City A has a higher mean annual precipitation (47.94 inches) compared to City M (43.89 inches).
- City A has a larger standard deviation (7.99 inches) compared to City M (5.60 inches), indicating greater variability in annual precipitation.