5.5 - Finding normal cut-offs and parameters

5.5 - Finding normal cut-offs and parameters

Sometimes the area under a normal curve is known and the measurement is missing. Work backwards from a probability to a standardised value, find cut-offs, and use one or two probability statements to determine an unknown mean or standard deviation.

From probability to a z-value

Measurements have mean 100100 and standard deviation 1212. Where does the highest 10%10\% begin? The boundary is above the mean, but its distance is determined by area, not by taking 10%10\% of the mean.

The supplied table gives Φ(z)=P(Z≤z)\Phi(z)=\mathrm P(Z\le z). To work backwards, convert the requested event into a left-tail probability first. Then find that probability in the body of the table and read the corresponding zz. The critical-values table supplies common results such as

Φ(1.282)≈0.90,Φ(1.645)≈0.95.\Phi(1.282)\approx0.90,\qquad \Phi(1.645)\approx0.95.

For a lower-tail area below 0.50.5, use symmetry: a left area 0.100.10 has z≈−1.282z\approx-1.282. A small right tail, by contrast, gives a positive zz.

Returning from a z-value to a measurement

For X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), a boundary xx corresponding to standardised value zz satisfies

x−μσ=z⟺x=μ+zσ.\frac{x-\mu}{\sigma}=z \quad\Longleftrightarrow\quad x=\mu+z\sigma.

Not in the formula booklet: rearrange the standardisation relationship. Common normal critical values are supplied in the tables.

The two examples use the same model and equal tail areas on opposite sides.

Worked example 1: an upper cut-off

Question. X∼N(100,144)X\sim\mathrm N(100,144). Find cc such that P(X>c)=0.10\mathrm P(X>c)=0.10.

Plan. Convert to a left area of 0.900.90, look up zz, then return to the measurement scale.

P(X<c)=0.90c−10012≈1.282c≈100+1.282(12)=115.384\begin{aligned} \mathrm P(X<c)&=0.90\\ \frac{c-100}{12}&\approx1.282\\ c&\approx100+1.282(12)=115.384 \end{aligned}

Answer. c≈115c\approx115, to 33 significant figures.

Check. The unrounded boundary gives (115.384−100)/12=1.282(115.384-100)/12=1.282, whose right area is about 0.100.10. The cut-off is above the mean, as an upper 10%10\% boundary should be.

Change only which tail has area 0.100.10. The distance from the mean stays the same, but its sign changes.

Worked example 2: a lower cut-off

Question. For X∼N(100,144)X\sim\mathrm N(100,144), find dd such that P(X<d)=0.10\mathrm P(X<d)=0.10.

Plan. Reflect the positive critical value, then use d=μ+zσd=\mu+z\sigma.

d−10012≈−1.282d≈100−1.282(12)=84.616\begin{aligned} \frac{d-100}{12}&\approx-1.282\\ d&\approx100-1.282(12)=84.616 \end{aligned}

Answer. d≈84.6d\approx84.6.

Check. The unrounded values obey c+d=200=2μc+d=200=2\mu, confirming that the two equal-area tails are reflections about 100100.

[DIAGRAM: asset_name: Equal tails and inverse cut-offs; asset_slug: normal-inverse-cutoffs; caption: Equal ten-percent tails have cut-offs equally far below and above the mean; description: A normal curve with mean 100 and standard deviation 12 has a shaded left tail of area 0.10 ending at 84.616 and a shaded right tail of area 0.10 starting at 115.384. The centre is labelled 100 and the corresponding z values minus 1.282 and 1.282 appear beneath the boundaries.]
Equal ten-percent tails have cut-offs equally far below and above the mean

For a probability absent from the critical-values table, search the main table. For example, a left area 0.840.84 lies between Φ(0.99)=0.8389\Phi(0.99)=0.8389 and Φ(1.00)=0.8413\Phi(1.00)=0.8413, so zz is between 0.990.99 and 1.001.00, approximately 0.9940.994 by interpolation. Retain enough precision until the final answer.

Watch out

A right-tail probability of 0.100.10 does not mean z=0.10z=0.10, and it does not mean Φ(z)=0.10\Phi(z)=0.10. Shade the requested tail, convert to a left area, then look up zz.

Find a different upper percentile in this check.

Finding an unknown mean or spread

A probability statement locates a boundary in standard-deviation units even when the original normal parameters are unknown. Translate that statement to (x−μ)/σ=z(x-\mu)/\sigma=z, then solve for the missing parameter. The standard deviation must be positive.

Worked example 3: finding a mean

Question. XX is normally distributed with unknown mean μ\mu and standard deviation 44. Given P(X<30)=0.90\mathrm P(X<30)=0.90, find μ\mu.

Plan. A left area of 0.900.90 corresponds to z≈1.282z\approx1.282; substitute the known cut-off and spread.

30−μ4≈1.28230−μ≈5.128μ≈24.872\begin{aligned} \frac{30-\mu}{4}&\approx1.282\\ 30-\mu&\approx5.128\\ \mu&\approx24.872 \end{aligned}

Answer. μ≈24.9\mu\approx24.9.

Check. The unrounded mean gives (30−24.872)/4=1.282(30-24.872)/4=1.282. The boundary 3030 is above the mean, consistent with 90%90\% of observations lying below it.

Keep the cut-off and probability, but now make the mean known and the spread unknown. The table lookup is unchanged; the algebra changes.

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