the vetrone family members are all cincinnati reds baseball fans. they went to five games last season. the…

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points. Here, $n = 5$, $x_1=266$, $x_2 = 201$, $x_3=197$, $x_4=188$, $x_5=162$. $\bar{x}=\frac{266 + 201+197+188+162}{5}=\frac{1014}{5}=202.8$

Step2: Calculate the range

The range is the difference between the maximum and minimum values. The maximum value is $266$ and the minimum value is $162$. Range $=266 - 162=104$

Step3: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$. $(x_1-\bar{x})=(266 - 202.8)=63.2$, $(x_2-\bar{x})=(201 - 202.8)=-1.8$, $(x_3-\bar{x})=(197 - 202.8)=-5.8$, $(x_4-\bar{x})=(188 - 202.8)=-14.8$, $(x_5-\bar{x})=(162 - 202.8)=-40.8$. $(x_1-\bar{x})^2=63.2^2 = 3994.24$, $(x_2-\bar{x})^2=(-1.8)^2 = 3.24$, $(x_3-\bar{x})^2=(-5.8)^2 = 33.64$, $(x_4-\bar{x})^2=(-14.8)^2 = 219.04$, $(x_5-\bar{x})^2=(-40.8)^2 = 1664.64$. $\sum_{i = 1}^{5}(x_{i}-\bar{x})^{2}=3994.24+3.24 + 33.64+219.04+1664.64=5914.8$. $s^{2}=\frac{5914.8}{4}=1478.7$

Step4: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$. $s=\sqrt{1478.7}\approx38.45$

Answer:

1. Mean cost per game: $202.8$
2. Range: $104$
3. Variance: $1478.7$
4. Standard deviation: $\approx38.45$