Question

mariana is considering a vacation and has obtained high - temperature data from the last two weeks for miami and los angeles.
miami: 76, 75, 83, 81, 83, 85
los angeles: 74, 63, 65, 62, 62, 72
which location has less variability in temperatures? explain how you arrived at your answer.

Explanation:

Step1: Calculate mean for Miami

Let \(x_{1 - 6}\) be Miami temperatures \(76,75,83,81,83,85\). Mean \(\bar{x}_{M}=\frac{76 + 75+83+81+83+85}{6}=\frac{483}{6} = 80.5\)

Step2: Calculate squared - differences for Miami

\((76 - 80.5)^2=(- 4.5)^2 = 20.25\), \((75 - 80.5)^2=(-5.5)^2 = 30.25\), \((83 - 80.5)^2=(2.5)^2 = 6.25\), \((81 - 80.5)^2=(0.5)^2 = 0.25\), \((83 - 80.5)^2=(2.5)^2 = 6.25\), \((85 - 80.5)^2=(4.5)^2 = 20.25\)
Sum of squared - differences \(S_{M}=20.25+30.25 + 6.25+0.25+6.25+20.25=83.5\)
Variance \(s_{M}^{2}=\frac{S_{M}}{n - 1}=\frac{83.5}{5}=16.7\)
Standard deviation \(s_{M}=\sqrt{16.7}\approx4.09\)

Step3: Calculate mean for Los Angeles

Let \(y_{1 - 6}\) be Los Angeles temperatures \(74,63,65,62,62,72\). Mean \(\bar{x}_{L}=\frac{74 + 63+65+62+62+72}{6}=\frac{398}{6}\approx66.33\)

Step4: Calculate squared - differences for Los Angeles

\((74 - 66.33)^2=(7.67)^2\approx58.83\), \((63 - 66.33)^2=(-3.33)^2\approx11.09\), \((65 - 66.33)^2=(-1.33)^2\approx1.77\), \((62 - 66.33)^2=(-4.33)^2\approx18.75\), \((62 - 66.33)^2=(-4.33)^2\approx18.75\), \((72 - 66.33)^2=(5.67)^2\approx32.15\)
Sum of squared - differences \(S_{L}=58.83+11.09+1.77+18.75+18.75+32.15 = 141.34\)
Variance \(s_{L}^{2}=\frac{S_{L}}{n - 1}=\frac{141.34}{5}=28.27\)
Standard deviation \(s_{L}=\sqrt{28.27}\approx5.32\)

Step5: Compare standard deviations

Since \(s_{M}\approx4.09\) and \(s_{L}\approx5.32\), Miami has less variability.

Answer:

Miami has less variability in temperatures because its standard deviation (\(s_{M}\approx4.09\)) is less than that of Los Angeles (\(s_{L}\approx5.32\)).