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.
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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.