The difference between the standard deviation of a population comparing it to the calculation of the standard deviation of a sample is that we divide by the population side n instead of (n-1).

We have to start by calculating the mean of the population first:

$\begin{gathered} \mu=\dfrac{1}{n}\sum ^n_{i=1}\, x_i \\ \mu=\dfrac{1}{6}(37+38+39+40+39+11) \\ \mu=\dfrac{204}{6} \\ \mu=34 \end{gathered}$

Now, we can calculate the standard deviation as:

$\sigma=\sqrt[]{\dfrac{1}{n}\sum^n_{i=1}\, (x_i-\mu)^2}$ $\begin{gathered} \sigma=\sqrt[]{\dfrac{1}{6}((37-34)^2+(38-34)^2+(39-34)^2+(40-34)^2+(39-34)^2+(11-34)^2)} \\ \sigma=\sqrt[]{\frac{1}{6}(3^2+4^2+5^2+6^2+5^2+(-23)^2)} \\ \sigma=\sqrt[]{\frac{1}{6}(9+16+25+36+25+529)} \\ \sigma=\sqrt[]{\frac{1}{6}(640)} \\ \sigma\approx\sqrt[]{106.67} \\ \sigma\approx10.33 \end{gathered}$

Answer: the standard deviation of this population is approximately 10.33

3 0

2 years ago

A caterer charges $120 for 15 people to cater a party and $7 per person after that.

Nitella [24]

Answer

120=7x+15

Step-by-step explanation:

7 0

4 years ago