- #1
jaejoon89
- 187
- 0
(to find distribution of sample mean)
Given
P((X1 - μ) / σ/√n) < Z < (X2 - μ) / σ/√n)) = P(a < Z < b) = phi(b) - phi(a)
where phi(z) = 1/sqrt(2*pi) * integral of exp(-z^2 / 2) dz from negative infinity to z
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I'm sure there's some statistical way of doing this with a TI 89, but how? The Normal cdf asks me for bounds, which I don't see what they would be here. so I figure that is not the correct function on the calculator. Using the calculator would be helpful since it's obviously not easy to solve this integral analytically.
Given
P((X1 - μ) / σ/√n) < Z < (X2 - μ) / σ/√n)) = P(a < Z < b) = phi(b) - phi(a)
where phi(z) = 1/sqrt(2*pi) * integral of exp(-z^2 / 2) dz from negative infinity to z
---
I'm sure there's some statistical way of doing this with a TI 89, but how? The Normal cdf asks me for bounds, which I don't see what they would be here. so I figure that is not the correct function on the calculator. Using the calculator would be helpful since it's obviously not easy to solve this integral analytically.