Question

Given that z is a standard normal random variable, find z for each situation. The area to the right of z is .1314. (Round to two decimal places) Answer The area to the left of z is .6700. (Round to two decimal places) Answer .44 The area to the right of z is .01. (Round to two decimal places) Answer

Answer 1

Answer:

$P(Z>0.1314) = 1-P(Z<0.1314) = 1-0.552=0.45$

We can use the following excel code: "=1-NORM.DIST(0.1314,0,1,TRUE)"  or the standard normal table

$P(Z<0.67) =0.75$

We can use the following excel code: "=NORM.DIST(0.75,0,1,TRUE)"  or the standard normal table.

$P(Z>0.01) = 1-P(Z<0.01) = 1-0.504=0.50$

We can use the following excel code: "=1-NORM.DIST(0.01,0,1,TRUE)"  or the standard normal table

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

$P(Z>0.1314) = 1-P(Z<0.1314) = 1-0.552=0.45$

We can use the following excel code: "=1-NORM.DIST(0.1314,0,1,TRUE)"  or the standard normal table

$P(Z<0.67) =0.75$

We can use the following excel code: "=NORM.DIST(0.75,0,1,TRUE)"  or the standard normal table.

$P(Z>0.01) = 1-P(Z<0.01) = 1-0.504=0.50$

We can use the following excel code: "=1-NORM.DIST(0.01,0,1,TRUE)"  or the standard normal table