Question

below are the jersey numbers of 11 players randomly selected from a football team. find the range, variance, and standard deviation for the given sample data. what do the results tell us? 75 68 40 50 41 48 62 19 79 33 71 range = 60 (round to one decimal place as needed.) sample standard deviation = (round to one decimal place as needed.)

Explanation:

Step1: Find the range

The range is the difference between the maximum and minimum values in the data - set.
The data set is \(75,68,40,50,41,48,62,19,79,33,71\).
The maximum value \(x_{max}=79\) and the minimum value \(x_{min}=19\).
Range \(=x_{max}-x_{min}=79 - 19=60\)

Step2: Calculate the mean

The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\), where \(n = 11\) and \(x_{i}\) are the data - points.
\(\sum_{i=1}^{11}x_{i}=75 + 68+40+50+41+48+62+19+79+33+71=586\)
\(\bar{x}=\frac{586}{11}\approx53.3\)

Step3: Calculate the squared differences

\((x_{1}-\bar{x})^2=(75 - 53.3)^2=(21.7)^2 = 470.89\)
\((x_{2}-\bar{x})^2=(68 - 53.3)^2=(14.7)^2 = 216.09\)
\((x_{3}-\bar{x})^2=(40 - 53.3)^2=(- 13.3)^2 = 176.89\)
\((x_{4}-\bar{x})^2=(50 - 53.3)^2=(-3.3)^2 = 10.89\)
\((x_{5}-\bar{x})^2=(41 - 53.3)^2=(-12.3)^2 = 151.29\)
\((x_{6}-\bar{x})^2=(48 - 53.3)^2=(-5.3)^2 = 28.09\)
\((x_{7}-\bar{x})^2=(62 - 53.3)^2=(8.7)^2 = 75.69\)
\((x_{8}-\bar{x})^2=(19 - 53.3)^2=(-34.3)^2 = 1176.49\)
\((x_{9}-\bar{x})^2=(79 - 53.3)^2=(25.7)^2 = 660.49\)
\((x_{10}-\bar{x})^2=(33 - 53.3)^2=(-20.3)^2 = 412.09\)
\((x_{11}-\bar{x})^2=(71 - 53.3)^2=(17.7)^2 = 313.29\)

Step4: Calculate the variance

The sample variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^2}{n - 1}\)
\(\sum_{i = 1}^{11}(x_{i}-\bar{x})^2=470.89+216.09+176.89+10.89+151.29+28.09+75.69+1176.49+660.49+412.09+313.29 = 3611.21\)
\(s^{2}=\frac{3611.21}{10}=361.121\)

Step5: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^{2}}\)
\(s=\sqrt{361.121}\approx19.0\)

Answer:

19.0