Respuesta :
Answer:
The result for this case would be [tex] z_{\alpha/2} =1.96[/tex]
And we can verify that:
[tex] P(-1.96 <Z< 1.96) = 0.95[/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
For this case we need to satisfy the following condition:
[tex] P(-z_{\alpha/2} < Z < z_{\alpha/2}) = 0.95[/tex]
Since the normal standard distribution is symmetric and the total area below the curve is 1 since we have a probability distribution, we can rewrite this expression like this
[tex] P(-z_{\alpha/2} < Z < z_{\alpha/2}) = 1-2P(Z<-z_{\alpha/2}) = 0.95[/tex]
So we can rewrite the last expression like this:
[tex] P(Z< -z_{\alpha/2}) = \frac{1-0.95}{2}= 0.025[/tex]
And we need to see on the normal standard distribution which value accumulates 0.025 of the area on the left tail. We can use the following excel code for example:
"=NORM.INV(0.025,0,1)"
And the result for this case would be [tex] z_{\alpha/2} =1.96[/tex]
And we can verify that:
[tex] P(-1.96 <Z< 1.96) = 0.95[/tex]
