Question

the vetrone family members are all cincinnati reds baseball fans. they went to five games last season. the cost of each game, including parking, tickets, and food, is listed below. $266, $201, $197, $188, $162. 1. what is the mean cost per game attended? 2. what is the range? 3. what is the variance? 4. what is the standard deviation?

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: $\approx38.45$