Question

listed below are the amounts (dollars) it costs for marriage proposal packages at different baseball stadiums. find the range, variance, and standard deviation for the given sample data. include appropriate units in the results. are there any outliers, and are they likely to have much of an effect on the measures of variation? 38 45 50 65 70 95 95 120 175 204 270 325 500 1750 2500 the range of the sample data is 2462 dollars (type an integer or a decimal. do not round.) the standard deviation of the sample data is 717.8 dollars (round to one decimal place as needed.) the variance of the sample data is 515236.8 dollars² (round to one decimal place as needed.) are there any outliers and if so, are they likely to have much of an effect on the measures of variation? a. yes, the largest amounts are much higher than the rest of the data, and appear to be outliers. it is not likely that these are having a large effect on the measures of variation b. yes, the smallest amounts are much lower than the rest of the data, and appear to be outliers. it is not likely that these are having a large effect on the measures of variation c. yes, the largest amounts are much higher than the rest of the data, and appear to be outliers. it is likely that these are having a large effect on the measures of variation d. no, there are not any outliers.

Explanation:

Step1: Calculate the range

The range is the difference between the maximum and minimum values. Given data: 38, 45, 50, 65, 70, 95, 95, 120, 175, 204, 270, 325, 500, 1750, 2500. The maximum value is 2500 and the minimum value is 38. So, range = 2500 - 38 = 2462 dollars.

Step2: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 15$ and $\sum_{i=1}^{15}x_{i}=38 + 45+50+65+70+95+95+120+175+204+270+325+500+1750+2500=6303$. So, $\bar{x}=\frac{6303}{15}=420.2$.

Step3: Calculate the variance

The sample - variance formula is $s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}$.
$(38-420.2)^{2}=(-382.2)^{2}=146076.84$, $(45 - 420.2)^{2}=(-375.2)^{2}=140775.04$, $(50-420.2)^{2}=(-370.2)^{2}=137048.04$, $(65 - 420.2)^{2}=(-355.2)^{2}=126167.04$, $(70-420.2)^{2}=(-350.2)^{2}=122640.04$, $(95-420.2)^{2}=(-325.2)^{2}=105755.04$, $(95-420.2)^{2}=(-325.2)^{2}=105755.04$, $(120-420.2)^{2}=(-300.2)^{2}=90120.04$, $(175-420.2)^{2}=(-245.2)^{2}=60123.04$, $(204-420.2)^{2}=(-216.2)^{2}=46742.44$, $(270-420.2)^{2}=(-150.2)^{2}=22560.04$, $(325-420.2)^{2}=(-95.2)^{2}=9063.04$, $(500-420.2)^{2}=(79.8)^{2}=6368.04$, $(1750-420.2)^{2}=(1329.8)^{2}=1768268.04$, $(2500-420.2)^{2}=(2079.8)^{2}=4325568.04$.
$\sum_{i = 1}^{15}(x_{i}-\bar{x})^{2}=7213315.2$.
$s^{2}=\frac{7213315.2}{14}\approx515236.8$ dollars$^{2}$.

Step4: Calculate the standard deviation

The sample - standard deviation $s=\sqrt{s^{2}}$. So, $s=\sqrt{515236.8}\approx717.8$ dollars.

Step5: Identify outliers and their effect

The values 1750 and 2500 are much higher than the rest of the data and are outliers. Outliers have a large effect on the measures of variation such as variance and standard - deviation.

Answer:

The range of the sample data is 2462 dollars.
The standard deviation of the sample data is 717.8 dollars.
The variance of the sample data is 515236.8 dollars$^{2}$.
C. Yes, the largest amounts are much higher than the rest of the data, and appear to be outliers. It is likely that these are having a large effect on the measures of variation.