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.
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.
1 answer