SOLUTION: If Z has a mean of 0 and a standard deviation of 1, determine the probability, P(|Z| <2.0) the answer is 0.9544 but i have no idea how they got that.

Algebra ->  Probability-and-statistics -> SOLUTION: If Z has a mean of 0 and a standard deviation of 1, determine the probability, P(|Z| <2.0) the answer is 0.9544 but i have no idea how they got that.      Log On


   



Question 1202329: If Z has a mean of 0 and a standard deviation of 1, determine the probability,
P(|Z| <2.0)

the answer is 0.9544 but i have no idea how they got that.

Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

You can use a stats calculator such as a TI84
https://www.statology.org/normal-probabilities-ti-84-calculator/
or spreadsheet software
https://www.statology.org/normalcdf-in-excel/
or something like this
https://davidmlane.com/normal.html

However, I'll be using a table instead.
This is the table I'll be using:
https://www.ztable.net/
A similar table is found in the back of any stats textbook.

Use that table to determine these two approximations:
P(Z < -2.00) = 0.02275
P(Z < 2.00) = 0.97725

Then we can compute the following.
P(a < Z < b) = P(Z < b) - P(Z < a)
P(-2 < Z < 2) = P(Z < 2) - P(Z < -2)
P(-2 < Z < 2) = 0.97725 - 0.02275
P(-2 < Z < 2) = 0.9545
P(|Z| < 2) = 0.9545

This is the approximate area under the Z curve between z = -2 and z = 2
The third link I mentioned at the top will draw out the diagram I'm describing.

The result 0.9545 is a bit off compared to 0.9544, but I think it's close enough.
This slight discrepancy is due to rounding error.