daily temperatures for two cities, salem and oxford, were recorded for one week. using statistics, answer…

daily temperatures for two cities, salem and oxford, were recorded for one week. using statistics, answer the question below! which city had the highest variability in temperatures? s m t w th f s salem 56 61 55 62 58 61 60 oxford 71 52 58 50 89 60 61 salem oxford both had the same variability.

daily temperatures for two cities, salem and oxford, were recorded for one week. using statistics, answer the question below! which city had the highest variability in temperatures? s m t w th f s salem 56 61 55 62 58 61 60 oxford 71 52 58 50 89 60 61 salem oxford both had the same variability.

Answer

Explanation:

Step1: Recall variance formula

Variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}$, where $x_{i}$ are data - points, $\bar{x}$ is the mean and $n$ is the number of data - points. First, find the mean for Salem. $\bar{x}_{Salem}=\frac{56 + 61+55+62+58+61+60}{7}=\frac{413}{7}=59$

Step2: Calculate squared - differences for Salem

$(56 - 59)^{2}=(-3)^{2}=9$, $(61 - 59)^{2}=2^{2}=4$, $(55 - 59)^{2}=(-4)^{2}=16$, $(62 - 59)^{2}=3^{2}=9$, $(58 - 59)^{2}=(-1)^{2}=1$, $(61 - 59)^{2}=2^{2}=4$, $(60 - 59)^{2}=1^{2}=1$ $\sum_{i = 1}^{7}(x_{i}-\bar{x}{Salem})^{2}=9 + 4+16+9+1+4+1=44$ $Var{Salem}=\frac{44}{7}\approx6.29$

Step3: Find the mean for Oxford

$\bar{x}_{Oxford}=\frac{71+52+58+50+89+60+61}{7}=\frac{441}{7}=63$

Step4: Calculate squared - differences for Oxford

$(71 - 63)^{2}=8^{2}=64$, $(52 - 63)^{2}=(-11)^{2}=121$, $(58 - 63)^{2}=(-5)^{2}=25$, $(50 - 63)^{2}=(-13)^{2}=169$, $(89 - 63)^{2}=26^{2}=676$, $(60 - 63)^{2}=(-3)^{2}=9$, $(61 - 63)^{2}=(-2)^{2}=4$ $\sum_{i = 1}^{7}(x_{i}-\bar{x}{Oxford})^{2}=64 + 121+25+169+676+9+4=1068$ $Var{Oxford}=\frac{1068}{7}\approx152.57$ Since $Var_{Oxford}>Var_{Salem}$, Oxford has higher variability.

Answer:

Oxford