Question

Assuming that these weights constitute an entire population, find the standard deviation of the population. ROUND YOUR ANSWER TO TWO DECIMAL PLACES

Answer 1

Given that the dataset constitutes a population, we will need to use the two formulas below

`\begin{gathered} \mu=(1)/(N)\sum_ix_i\rightarrow\text{ population mean} \\ \sigma=(√(\sum_i(x_i-\mu)^2))/(√(N))\rightarrow\text{ population standard deviation} \\ \end{gathered}`

Thus, in our case,

`\begin{gathered} \Rightarrow\mu=(1)/(5)(93+97+111+95+109)=(505)/(5)=101 \\ \Rightarrow\mu=101 \end{gathered}`

Finding the standard deviation,

`\begin{gathered} \Rightarrow\sigma=(√((93-101)^2+(97-101)^2+...+(109-101)^2))/(√(5)) \\ \Rightarrow\sigma=\sqrt{(280)/(5)}=√(56)\approx7.48 \end{gathered}`

Thus, the standard deviation of the population is 7.48