Question

analyzing average wind patterns
the monthly wind speeds over a one - year period at denver international airport were recorded and the values for each month averaged. the average monthly wind speeds, in mph, from january to december during that time period were 9.7, 10.0, 10.8, 11.9, 11.0, 10.7, 10.3, 10.1, 9.9, 9.9, 9.6, and 10.1
use the statistics calculator to find the statistical measures
the median of the data set is
the mean of the data set is
the population standard deviation of the data set is

Explanation:

Step1: Arrange data in ascending order

9.6, 9.7, 9.9, 9.9, 10.0, 10.1, 10.1, 10.3, 10.7, 10.8, 11.0, 11.9

Step2: Calculate the median

Since there are 12 data - points (an even number), the median is the average of the 6th and 7th ordered values. Median=$\frac{10.1 + 10.1}{2}=10.1$

Step3: Calculate the mean

Mean=$\frac{9.6+9.7 + 9.9+9.9+10.0+10.1+10.1+10.3+10.7+10.8+11.0+11.9}{12}=\frac{123}{12}=10.25$

Step4: Calculate the population standard deviation

First, calculate the variance. The formula for population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}$, where $x_{i}$ are the data - points, $\mu$ is the mean, and $n$ is the number of data - points.
$(9.6 - 10.25)^{2}=(-0.65)^{2}=0.4225$, $(9.7 - 10.25)^{2}=(-0.55)^{2}=0.3025$, $(9.9 - 10.25)^{2}=(-0.35)^{2}=0.1225$ (twice), $(10.0 - 10.25)^{2}=(-0.25)^{2}=0.0625$, $(10.1 - 10.25)^{2}=(-0.15)^{2}=0.0225$ (twice), $(10.3 - 10.25)^{2}=(0.05)^{2}=0.0025$, $(10.7 - 10.25)^{2}=(0.45)^{2}=0.2025$, $(10.8 - 10.25)^{2}=(0.55)^{2}=0.3025$, $(11.0 - 10.25)^{2}=(0.75)^{2}=0.5625$, $(11.9 - 10.25)^{2}=(1.65)^{2}=2.7225$
$\sum_{i = 1}^{12}(x_{i}-10.25)^{2}=0.4225+0.3025 + 2\times0.1225+0.0625+2\times0.0225+0.0025+0.2025+0.3025+0.5625+2.7225 = 5.04$
Variance $\sigma^{2}=\frac{5.04}{12}=0.42$
Population standard deviation $\sigma=\sqrt{0.42}\approx0.648$

Answer:

Median: 10.1
Mean: 10.25
Population standard deviation: 0.648