Question

(a) which city had a smaller range of noon temperatures? (b) which city had noon temperatures with a larger interquartile range (iqr)? (c) which city had a larger median noon temperature? (a) which city had the highest noon temperature? select select select select

Explanation:

Step1: Find the highest noon temperature

The highest value (maximum) for City B is around 95°F and for City A is around 75°F. So City B has the highest noon temperature.

Step2: Find the city with smaller range

Range = Maximum - Minimum. For City B: \(95 - 70= 25\). For City A: \(75 - 65 = 10\). So City A has a smaller range.

Step3: Find the city with larger IQR

IQR = \(Q_3 - Q_1\). For City B: \(Q_3\approx90\), \(Q_1\approx88\), \(IQR = 90 - 88=2\). For City A: \(Q_3\approx82\), \(Q_1\approx85\) (wait no, correct: for box - and - whisker plot, the box represents \(Q_1\), median (\(Q_2\)), \(Q_3\). For City B: \(Q_1 = 88\), \(Q_3=90\), \(IQR = 90 - 88 = 2\). For City A: \(Q_1 = 85\), \(Q_3=80\) (wrong, actually for City A: \(Q_1 = 85\), \(Q_3=80\) no - looking at the plot again. Wait, no: the left end of the box is \(Q_1\), the right end is \(Q_3\). For City B: \(Q_1 = 88\), \(Q_3=90\), \(IQR=2\). For City A: \(Q_1 = 80\), \(Q_3=85\), \(IQR = 85 - 80=5\). So City A has a larger IQR.

Step4: Find the city with larger median

Median (\(Q_2\)): For City B, the line in the box is around 89. For City A, the line in the box is around 82. So City B has a larger median.

Answer:

(a) City A; (b) City A; (c) City B