Question

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

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

Step2: Calculate the median

Since $n = 12$ (even - numbered data set), median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+ 1)$th values. $\frac{n}{2}=6$ and $\frac{n}{2}+1 = 7$. The 6th value is $10.1$ and the 7th value is $10.1$. Median $=\frac{10.1 + 10.1}{2}=10.1$

Step3: Calculate the mean

Mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}=\frac{9.6+9.9\times3 + 10.0+10.1\times3+10.3+10.7+10.8+11.0+11.9}{12}=\frac{124}{12}\approx10.333$

Step4: Calculate the population standard - deviation

The formula for population standard deviation is $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}}$.
First, find $(x_{i}-\bar{x})^{2}$ for each $x_{i}$:
$(9.6 - 10.333)^{2}=(- 0.733)^{2}=0.537289$
$(9.9 - 10.333)^{2}=(-0.433)^{2}=0.187489$ (three times)
$(10.0 - 10.333)^{2}=(-0.333)^{2}=0.110889$
$(10.1 - 10.333)^{2}=(-0.233)^{2}=0.054289$ (three times)
$(10.3 - 10.333)^{2}=(-0.033)^{2}=0.001089$
$(10.7 - 10.333)^{2}=(0.367)^{2}=0.134689$
$(10.8 - 10.333)^{2}=(0.467)^{2}=0.218089$
$(11.0 - 10.333)^{2}=(0.667)^{2}=0.444889$
$(11.9 - 10.333)^{2}=(1.567)^{2}=2.455489$

$\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}=0.537289+3\times0.187489 + 0.110889+3\times0.054289+0.001089+0.134689+0.218089+0.444889+2.455489 = 4.74$
$\sigma=\sqrt{\frac{4.74}{12}}\approx0.632$

Answer:

The median of the data set is $10.1$
The mean of the data set is $10.333$
The population standard deviation of the data set is $0.632$