Inverse normal goes from area to value: The inverse normal does the reverse of normalcdf: given a left-tail probability P(X < x), it returns the value x.
On the GDC use invNorm(area, μ, σ).
Inverse normal: given the area (probability), find the cut-off value back on the axis.
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IB-style question — find the value
X ~ N(50, 10²).
Find the value x such that P(X < x) = 0.9.
Step by step
- Inverse normal with the left-tail area.
- Read the result.
Final answer
x ≈ 62.8.
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Convert to a left-tail area first: invNorm always uses the left area.
For P(X > x) = p, use the left area 1 − p.
For a symmetric middle proportion, split the leftover equally into two tails.
IB-style question — top tail
X ~ N(50, 10²).
Find x such that P(X > x) = 0.05.
Step by step
- Convert to a left area.
- Inverse normal.
Final answer
x ≈ 66.4.
Middle 90% → tails of 0.05: For the central 90%, each tail is 0.05; the boundaries are invNorm(0.05,…) and invNorm(0.95,…).
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Get z from invNorm, then σ = (x − μ)/z: If σ is unknown, find the z-value from the probability with invNorm(area, 0, 1), then rearrange z = (x − μ)/σ to get σ = (x − μ)/z.
IB-style question — find σ
X ~ N(100, σ²) and P(X < 118) = 0.8.
Find σ.
Step by step
- z-value for a left area of 0.8.
- Rearrange z = (x − μ)/σ.
Final answer
σ ≈ 21.4.
Use the standardised z: To find σ you must go through z (invNorm with μ = 0, σ = 1) — you can't use invNorm with the unknown σ directly.
Get z, then μ = x − zσ: If the mean is unknown, find z from the probability with invNorm(area, 0, 1), then rearrange z = (x − μ)/σ to μ = x − zσ.
IB-style question — find μ
X ~ N(μ, 5²) and P(X < 20) = 0.1.
Find μ.
Step by step
- z-value for a left area of 0.1.
- Rearrange to μ = x − zσ.
Final answer
μ ≈ 26.4.
Mind the sign of z: Here z is negative (0.1 is a left tail), so μ = x − zσ adds a positive amount — 20 is below the mean.