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

$Z=\frac{x-\mu}{\sigma}$

Where

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

So, replacing you have

$Z=\frac{x-\mu}{\sigma}=\frac{50-58}{12}=-\frac{8}{12}$

Simplifying you have

$Z=-\frac{8}{12}=-\frac{2}{3}=-0.67$