Question

You have a normal distribution with a mean of 58 and a standard deviation of 12. Aparticular value of x is 50. What is the z-score of 50 given that normal distributions?Users

Answer 1

To do this you can use the following formula that is used to standardize a random variable which has a normal distribution

`Z=(x-\mu)/(\sigma)`

Where

`\begin{gathered} \mu\text{ represents the mean and, } \\ \sigma\text{ represents the standard desviation} \end{gathered}`

So, replacing you have

`Z=(x-\mu)/(\sigma)=(50-58)/(12)=-(8)/(12)`

Simplifying you have

`Z=-(8)/(12)=-(2)/(3)=-0.67`