Question

About 8% of the population has a particular genetic mutation. 900 people are randomly selected.Find the standard deviation for the number of people with the genetic mutation in such groups of 900. Round your answer to two decimal places.

Answer 1

To find the standard deviation, the formula is

`\begin{gathered} \sigma=√(npq) \\ n\text{ is number selected } \\ p\text{ is sucessful trials} \\ q\text{ is failed trials} \end{gathered}`

Given

`\begin{gathered} p=(8)/(100)=0.08 \\ q=1-0.08=0.92 \\ n=900 \end{gathered}`

Substitute the values of the variables into the formula above

`\begin{gathered} \sigma=√(npq) \\ \sigma=√(900*0.08*0.92) \\ \sigma=√(66.24) \\ \sigma=8.14\text{ \lparen two decimal places\rparen} \end{gathered}`

Hence, the answer is 8.14 (two decimal places)