Question

the numbers of regular - season wins for 10 football teams in a given season are given below. determine the range, mean, variance, and standard deviation of the population data set. 2, 8, 15, 5, 14, 8, 11, 9, 4, 6. the range is (simplify your answer.)

Explanation:

Step1: Arrange data in ascending order

$2,4,5,6,8,9,11,14,15,15$

Step2: Calculate the range

Range = Max - Min = $15 - 2=13$

Step3: Calculate the mean

$\bar{x}=\frac{2 + 4+5+6+8+9+11+14+15+15}{10}=\frac{89}{10}=8.9$

Step4: Calculate the variance

$\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}$
$(2 - 8.9)^{2}=(-6.9)^{2}=47.61$
$(4 - 8.9)^{2}=(-4.9)^{2}=24.01$
$(5 - 8.9)^{2}=(-3.9)^{2}=15.21$
$(6 - 8.9)^{2}=(-2.9)^{2}=8.41$
$(8 - 8.9)^{2}=(-0.9)^{2}=0.81$
$(9 - 8.9)^{2}=(0.1)^{2}=0.01$
$(11 - 8.9)^{2}=(2.1)^{2}=4.41$
$(14 - 8.9)^{2}=(5.1)^{2}=26.01$
$(15 - 8.9)^{2}=(6.1)^{2}=37.21$
$(15 - 8.9)^{2}=(6.1)^{2}=37.21$
$\sum_{i = 1}^{10}(x_{i}-8.9)^{2}=47.61+24.01+15.21+8.41+0.81+0.01+4.41+26.01+37.21+37.21 = 190.9$
$\sigma^{2}=\frac{190.9}{10}=19.09$

Step5: Calculate the standard - deviation

$\sigma=\sqrt{\sigma^{2}}=\sqrt{19.09}\approx4.37$

Answer:

Range: $13$
Mean: $8.9$
Variance: $19.09$
Standard deviation: $\approx4.37$