Question

Define z_alpha to be a z-score with an area of alpha to the right. For Example: z_0.10 means P(Z > z_0.10) = 0.10. We would also have, for example; P(Z < -z_0.10) = 0.10 because of the symmetric property of the normal distribution. Questions: (a)Find P(-z_0.025 < Z < z_0.025) (b)Find P(-z_alpha/2 < Z < 2_alpha/2). (c)Find the value of z_alpha (rounded to 3 decimals) such that P(Z > z_alpha) = 0.05.

Answer 1

Answer:

a) P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :$z_{crit}= \pm 1.96$

b) P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

c) For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$

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

Part a

P(-z_0.025 < Z < z_0.025)

For this case we want a quantile that accumulates 0.025 of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(0.025,0,1)"

"=NORM.INV(0.025,0,1)"

And for this case the two values are :$z_{crit}= \pm 1.96$

Part b

P(-z_{\alpha/2} < Z < z_{\alpha/2})

For this case we want a quantile that accumulates $\alpha/2$ of the area on the tails of the normal standard distribution, and for this case we can calculate the z value with the following excel codes:

"=NORM.INV(alpha/2,0,1)"

"=NORM.INV(alpha/2,0,1)"

Part c

For this case we want to find a value of z that satisfy:

P(Z > z_alpha) = 0.05.

And we can use the following excel code:

"=NORM.INV(0.95,0,1)"

And we got $z_{\alpha/2}=1.64$