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

Respuesta :

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

[tex]Z=\frac{x-\mu}{\sigma}[/tex]

Where

[tex]\begin{gathered} \mu\text{ represents the mean and, } \\ \sigma\text{ represents the standard desviation} \end{gathered}[/tex]

So, replacing you have

[tex]Z=\frac{x-\mu}{\sigma}=\frac{50-58}{12}=-\frac{8}{12}[/tex]

Simplifying you have

[tex]Z=-\frac{8}{12}=-\frac{2}{3}=-0.67[/tex]