Respuesta :
Answer:
[tex] E(Z) =\frac{E(X -\mu)}{\sigma} = 0[/tex]
[tex] Var(Z) = Var(\frac{X -\mu}{\sigma}) = \frac{Var(X-\mu}{\sigma^2}= \frac{\sigma^2}{\sigma^2}=1[/tex]
[tex]Sd(Z)= \sqrt{Var(Z)} = \sqrt{1} =1[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(75.3,20.15)[/tex]
Where [tex]\mu=75.3[/tex] and [tex]\sigma=20.15[/tex]
For this case if we transform this to a Z normal distribution the new variable will be defined as:
[tex] Z= \frac{X -\mu}{\sigma}[/tex]
The expected value fo Z is:
[tex] E(Z) =\frac{E(X -\mu)}{\sigma} = 0[/tex]
And the variance would be given by:
[tex] Var(Z) = Var(\frac{X -\mu}{\sigma}) = \frac{Var(X-\mu}{\sigma^2}= \frac{\sigma^2}{\sigma^2}=1[/tex]
[tex]Sd(Z)= \sqrt{Var(Z)} = \sqrt{1} =1[/tex]
