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Normal Probability
Your Discussion Posting: If X ~ N (750, 12), we can determine the P(x > 784) as follows:
μ = 750
σ = 12
x = 784
Z = x − μ σ
Z = 784 −750 12
Z =2.8333...
Z = 2.83
P ( X > 784 ) = 1 − 0.9977
P ( X > 784 ) = 0.0023
So, the probability of an observation greater than 784 with a standard deviation of 12 is 0.0023.
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Determine the P (x > 784) for each of the following two scenarios:
X ~ N (750, 24)doubling
X ~ N (750, 6)halving
For each scenario(1) (2)X ~ N (750, 12)
- The Show-Your-Work process P(x > 784)
- given value(s) appropriate symbol(s)
- with symbols only
- with numbers plugged in
- Unrounded
- Rounded
- A normal curve with the following:
- area of interest shaded
- labeled
- meansaxes
- X X
- ZZ
- AAA
- A detailed explanation of how the change in the standard deviation does or does not affect each of the following:
- center
- spread
- shape
- P(x > 784)
I got the first part but having trouble with X~N(750,6)