prof: mr. nougues\ncourse: probability and statistics\nconsider the data below. find the population variance…

Explanation:

Step1: Calculate the mean

First, sum the data values: $40 + 23+41+50+49+32+41+29+52+58 = 415$. There are $n = 10$ data - points. The mean $\mu=\frac{415}{10}=41.5$.

Step2: Calculate the squared - differences

For each data value $x_i$, calculate $(x_i-\mu)^2$. $(40 - 41.5)^2=(-1.5)^2 = 2.25$; $(23 - 41.5)^2=(-18.5)^2=342.25$; $(41 - 41.5)^2=(-0.5)^2 = 0.25$; $(50 - 41.5)^2=(8.5)^2 = 72.25$; $(49 - 41.5)^2=(7.5)^2 = 56.25$; $(32 - 41.5)^2=(-9.5)^2 = 90.25$; $(41 - 41.5)^2=(-0.5)^2 = 0.25$; $(29 - 41.5)^2=(-12.5)^2 = 156.25$; $(52 - 41.5)^2=(10.5)^2 = 110.25$; $(58 - 41.5)^2=(16.5)^2 = 272.25$.

Step3: Calculate the population variance

The population variance $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}$. $\sum_{i = 1}^{10}(x_i - 41.5)^2=2.25+342.25+0.25+72.25+56.25+90.25+0.25+156.25+110.25+272.25 = 1108$. So, $\sigma^{2}=\frac{1108}{10}=110.8$.

Step4: Calculate the population standard deviation

The population standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{110.8}\approx10.526$.

Answer:

Population variance: $110.8$, Population standard deviation: $\approx10.526$