S1 5.1 - Finding Normal parameters

S1 5.1 - Finding Normal parameters

Work backwards from a Normal probability to a boundary, mean or standard deviation. Translate percentile conditions into signed standardised values, then solve two conditions together when both parameters are unknown.

Inverse standard Normal probabilities

Where does the lowest 10%10\% of a Normal distribution end? Its boundary must lie below the mean, so its standardised value must be negative. Choosing the sign is as important as finding the table entry.

A percentile is a boundary below which a stated percentage of the distribution lies. For example, the 90th percentile has cumulative probability 0.900.90. It leaves probability 0.100.10 to its right.

The booklet has two useful tables. The cumulative table gives Φ(z)=P(Z≤z)\Phi(z)=\mathrm P(Z\le z). The percentage-points table gives nonnegative zz such that

P(Z>z)=p.\mathrm P(Z>z)=p.

For example, its entry p=0.1000p=0.1000, z=1.2816z=1.2816 means that 10%10\% lies above 1.28161.2816, not below it. Its entry p=0.1500p=0.1500, z=1.0364z=1.0364 identifies the 85th percentile. If the required cumulative probability is 0.300.30, use p=0.30p=0.30 and reflect the listed 0.52440.5244 to −0.5244-0.5244.

[DIAGRAM: asset_name: Opposite ten-percent tails; asset_slug: s1-normal-inverse-signs; caption: The lowest and highest ten percent have equal areas at opposite signed standardised boundaries.; description: A standard Normal curve shades the lower 10 percent to the left of negative 1.2816 and the upper 10 percent to the right of positive 1.2816. Both signed boundaries and zero are labelled; the central 80 percent is unshaded.]
The lowest and highest ten percent have equal areas at opposite signed standardised boundaries.

When the cumulative probability is supplied as a table entry, you can also read the cumulative table backwards: Φ(c)=0.9332\Phi(c)=0.9332 gives c=1.50c=1.50. No interpolation is required. A calculator inverse-Normal function usually asks for the cumulative left probability; check that convention before entering a right-tail probability.

Key idea

First locate the boundary relative to the mean. A cumulative probability below 0.50.5 has a negative standardised boundary; one above 0.50.5 has a positive boundary.

Use the full tabulated four decimal places for percentage points in subsequent equations. Rounding 1.03641.0364 prematurely to 1.041.04, for example, can alter a calculated standard deviation.

Worked example 1: an upper ten-percent boundary

For Z∼N(0,1)Z\sim\mathrm N(0,1), find cc such that P(Z>c)=0.10\mathrm P(Z>c)=0.10.

Plan. This is exactly the right-tail convention of the percentage-points table.

p=0.1000⟹c=1.2816.p=0.1000\quad\Longrightarrow\quad c=1.2816.

Answer. c=1.2816c=1.2816.

Check. Its cumulative left probability is 0.900.90, so the boundary is above zero and leaves the required 0.100.10 on its right.

Change only the side of the probability from right to left, keeping 10%10\%. Symmetry reverses the sign of the boundary.

Worked example 2: a lower ten-percent boundary

For Z∼N(0,1)Z\sim\mathrm N(0,1), find dd such that P(Z<d)=0.10\mathrm P(Z<d)=0.10.

Plan. Reflect the upper ten-percent boundary in zero.

P(Z<−1.2816)=P(Z>1.2816)=0.10.\mathrm P(Z<-1.2816)=\mathrm P(Z>1.2816)=0.10.

Answer. d=−1.2816d=-1.2816.

Check. The boundary lies below the mean and 90%90\% lies above it. A positive answer would instead leave more than half the distribution to its left.

Match the requested tail before reading off a sign.

Unknown mean or standard deviation

A probability condition tells you the standardised distance of a boundary from the mean. With one unknown parameter, this gives one equation to solve.

If a boundary xx corresponds to standardised value zz, then

x−μσ=z,x=μ+zσ.\frac{x-\mu}{\sigma}=z,\qquad x=\mu+z\sigma.

The second form reads naturally: boundary equals mean plus a signed number of standard deviations. To find a percentile boundary when both parameters are known, simply substitute into this form. For example, a positive zz adds a distance above the mean; a negative zz subtracts a distance.

Recovering a parameter

μ=x−zσ,σ=x−μz(z≠0).\mu=x-z\sigma,\qquad \sigma=\frac{x-\mu}{z}\quad(z\ne0).

The standard deviation must be positive. A condition at the median has z=0z=0: it gives x=μx=\mu but supplies no information about σ\sigma.

These are rearrangements of standardisation, not separately provided formulae.

Find the signed zz first. Do not treat a percentage or probability as the standardised value.

Worked example 3: finding a mean

X∼N(μ,52)X\sim\mathrm N(\mu,5^2) and P(X>61)=0.10\mathrm P(X>61)=0.10. Find μ\mu.

Plan. An upper tail of 0.100.10 puts the boundary 1.28161.2816 standard deviations above the mean.

61−μ5=1.2816,μ=61−5(1.2816)=54.592.\frac{61-\mu}{5}=1.2816, \qquad \mu=61-5(1.2816)=54.592.

Answer. μ=54.6\mu=54.6, to three significant figures.

Check. 54.592+1.2816(5)=6154.592+1.2816(5)=61. The boundary exceeds the mean, which agrees with the small upper-tail probability.

Keep the boundary and probability fixed, but change which parameter is known. Now the mean is given and the scale must be found.

Worked example 4: finding a standard deviation

Y∼N(55,σ2)Y\sim\mathrm N(55,\sigma^2) and P(Y>61)=0.10\mathrm P(Y>61)=0.10. Find σ\sigma.

Plan. Use the same positive percentage point, then divide the distance above the mean by it.

61−55σ=1.2816,σ=61.2816=4.681647….\frac{61-55}{\sigma}=1.2816, \qquad \sigma=\frac6{1.2816}=4.681647\ldots.

Answer. σ=4.68\sigma=4.68, to three significant figures.

Check. The result is positive, and 55+1.2816(4.681647…)=6155+1.2816(4.681647\ldots)=61. Dividing by σ2\sigma^2 instead would use the wrong scale.

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