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Worked solution

Find an unknown mean from a tail probability

If X∼N(μ,62)X\sim N(\mu,6^2) and P(X>82)=0.0478P(X>82)=0.0478, find μ.

CJC · Lecture · Example 5

Standardise the given tail

Use Z=(X−μ)/σZ=(X-\mu)/\sigma with σ6:

P ⁣(X−μσ>82−μσ)=0.0478,P ⁣(Z>82−μ6)=0.0478.P\!\left(\frac{X-\mu}{\sigma}>\frac{82-\mu}{\sigma}\right)=0.0478,\qquad P\!\left(Z>\frac{82-\mu}{6}\right)=0.0478.

The source’s invNorm wizard uses area 0.0478, μ0, σ1 and RIGHT. invNorm(0.0478,0,1,RIGHT) displays 1.666569689.

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Solve for the original mean

Match the standardised bound to that quantile:

82−μ6≈1.666569689⟹μ≈82−6(1.666569689)=72.000581866.\frac{82-\mu}{6}\approx 1.666569689\quad\Longrightarrow\quad\mu\approx 82-6(1.666569689)=72.000581866.

Therefore μ=72.0\boxed{\mu=72.0} (3 s.f.), as printed. The cutoff 82 is above the inferred mean, consistent with its small right-tail area.

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Normal standard deviation 6, unknown mean μ. The right-tail area beyond 82 is.0478; the standard-normal right cutoff is approximately 1.666569689. Thus 82−μ=6×1.666569689 and μ≈72.0. The cutoff 82 lies right of the mean.
Normal standard deviation 6, unknown mean μ. The right-tail area beyond 82 is.0478; the standard-normal right cutoff is approximately 1.666569689. Thus 82−μ=6×1.666569689 and μ≈72.0. The cutoff 82 lies right of the mean.
Normal standard deviation 6, unknown mean μ. The right-tail area beyond 82 is.0478; the standard-normal right cutoff is approximately 1.666569689. Thus 82−μ=6×1.666569689 and μ≈72.0. The cutoff 82 lies right of the mean.

Common mistakes

Use μ0 and σ1 when finding the standard-normal quantile, then solve the resulting equation for the original μ. Do not enter the unknown original mean into the calculator command.

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