Question

prof: mr. nougues
course: probability and statistics
consider the data below. find the population variance and standard deviation of it
40 23 41 50 49 32 41 29 52 58.

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 - point $x_i$, 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 $\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}$.
$\sum_{i = 1}^{10}(x_i-\mu)^2=2.25+342.25 + 0.25+72.25+56.25+90.25+0.25+156.25+110.25+272.25 = 1102$.
So, $\sigma^{2}=\frac{1102}{10}=110.2$.

Step4: Calculate the population standard deviation

The population standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{110.2}\approx10.5$.

Answer:

Population variance: $110.2$, Population standard deviation: $\approx10.5$