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? 33 6 80 3 32 75 98 10 71 19 9 range = 95.0 (round to one decimal place as needed.) sample standard deviation = (round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

First, find the sum of the data values: \(33 + 6+80 + 3+32 + 75+98 + 10+71 + 19+9=436\).
The number of data points \(n = 11\).
The mean \(\bar{x}=\frac{436}{11}\approx39.6\)

Step2: Calculate the squared differences

For each data value \(x_i\), calculate \((x_i-\bar{x})^2\):

• For \(x_1 = 33\): \((33 - 39.6)^2=(-6.6)^2 = 43.56\)
• For \(x_2 = 6\): \((6 - 39.6)^2=(-33.6)^2=1128.96\)
• For \(x_3 = 80\): \((80 - 39.6)^2=(40.4)^2 = 1632.16\)
• For \(x_4 = 3\): \((3 - 39.6)^2=(-36.6)^2 = 1339.56\)
• For \(x_5 = 32\): \((32 - 39.6)^2=(-7.6)^2 = 57.76\)
• For \(x_6 = 75\): \((75 - 39.6)^2=(35.4)^2 = 1253.16\)
• For \(x_7 = 98\): \((98 - 39.6)^2=(58.4)^2 = 3410.56\)
• For \(x_8 = 10\): \((10 - 39.6)^2=(-29.6)^2 = 87，6.16\)
• For \(x_9 = 71\): \((71 - 39.6)^2=(31.4)^2 = 985.96\)
• For \(x_{10}=19\): \((19 - 39.6)^2=(-20.6)^2 = 424.36\)
• For \(x_{11}=9\): \((9 - 39.6)^2=(-30.6)^2 = 936.36\)

Sum of squared differences \(\sum_{i = 1}^{n}(x_i-\bar{x})^2=43.56+1128.96 + 1632.16+1339.56+57.76+1253.16+3410.56+876.16+985.96+424.36+936.36 = 12188.6\)

Step3: Calculate the sample variance

The sample variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}=\frac{12188.6}{11-1}=\frac{12188.6}{10}=1218.9\)

Step4: Calculate the sample standard deviation

The sample standard deviation \(s=\sqrt{s^2}=\sqrt{1218.9}\approx34.9\)

Answer:

Sample standard deviation \(=34.9\)