Question

jillian collected data about the average daily temperatures for two cities. she organized her data in the given table.
average daily temperature (°f)

	jan	feb	mar	apr	may	jun	jul	aug	sept	oct	nov	dec
city b	30	52	58	67	72	80	85	88	82	75	55	35

click on all the true statements about the data.
a. over the course of the year, the mean temperature is the same for both cities.
b. over the course of the year, the mean temperature is greater in city a than city b.
c. over the course of the year, the mean temperature is greater in city b than city a.
d. over the course of the year, the range of temperatures is greater in city a than city b.
e. over the course of the year, the range of temperatures is greater in city b than city a.

Explanation:

Step1: Calculate mean of City A

The mean formula is $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For City A, $n = 12$, and $\sum_{i=1}^{12}x_{i}=2 + 12+30 + 38+55+62+85+90+78+60+35+11=568$. So the mean of City A, $\bar{x}_A=\frac{568}{12}\approx47.33$.

Step2: Calculate mean of City B

For City B, $\sum_{i = 1}^{12}x_{i}=30+52+58+67+72+80+85+88+82+75+55+35 = 789$. So the mean of City B, $\bar{x}_B=\frac{789}{12}=65.75$. Since $65.75>47.33$, the mean of City B is greater than that of City A.

Step3: Calculate range of City A

The range formula is $R=\text{Max}-\text{Min}$. For City A, $\text{Max}=90$, $\text{Min}=2$, so $R_A=90 - 2=88$.

Step4: Calculate range of City B

For City B, $\text{Max}=88$, $\text{Min}=30$, so $R_B=88 - 30 = 58$. Since $88>58$, the range of City A is greater than that of City B.

Answer:

C. Over the course of the year, the mean temperature is greater in City B than City A.
D. Over the course of the year, the range of temperatures is greater in City A than City B.