S1 5.1 - Normal probabilities

S1 5.1 - Normal probabilities

Use the shape, symmetry and parameters of a Normal distribution to describe continuous measurements. Standardise values and use the supplied cumulative tables to calculate probabilities in either tail or between two limits.

Shape symmetry mean and variance

Is a measurement of 8686 unusually large? That depends on the usual value and the spread: 8686 is far less surprising when measurements vary widely. A Normal model supplies both pieces of information.

Normal distribution

A Normal distribution is a continuous, bell-shaped probability distribution symmetric about its mean μ\mu. We write

X∼N(μ,σ2),σ>0,X\sim\mathrm N(\mu,\sigma^2),\qquad \sigma>0,

where σ2\sigma^2 is the variance and σ\sigma is the standard deviation.

The mean, median and mode coincide at the centre. Increasing σ\sigma spreads the curve out and lowers its peak, while keeping the total area equal to 11. Half the area lies on each side of μ\mu.

[DIAGRAM: asset_name: Normal centre and spread; asset_slug: s1-normal-centre-spread; caption: Both Normal curves have mean 50 and total area one. The curve with standard deviation 10 is more spread out.; description: Exact Normal curves with mean 50 and standard deviations 5 and 10, sharing a horizontal measurement axis. Their peaks align at 50. A dashed central line marks the mean, and two side labels identify one half of each distribution on either side.]
Both Normal curves have mean 50 and total area one. The curve with standard deviation 10 is more spread out.

Probabilities are areas under the curve, not heights. A single exact value has zero area, so P(X=c)=0\mathrm P(X=c)=0. Thus P(X<c)=P(X≤c)\mathrm P(X<c)=\mathrm P(X\le c), and including or excluding interval endpoints does not change a Normal probability. This differs from a discrete distribution, where one supported value can have positive probability.

A Normal model allows values across the whole real line. For a physical quantity that cannot be negative, it is an approximation that may be useful when the impossible tail is negligible. Symmetry and a single central peak support the model; pronounced skewness or several peaks count against it. Being continuous alone is insufficient.

Watch out

In X∼N(40,9)X\sim\mathrm N(40,9), the standard deviation is 9=3\sqrt9=3, not 99. The second parameter is the variance.

Read both the parameters and the meaning of a continuous probability.

Standardising a value

The useful question is how many standard deviations a measurement lies above or below its mean. Subtracting the mean removes its location; dividing by the standard deviation measures the distance in standard-deviation units.

For a model with mean 8080 and standard deviation 44, the value 8686 is 66 units above the mean, or 6/4=1.56/4=1.5 standard deviations above it. The value 7474 is the same distance below it.

Standardising a Normal variable

If X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), then

Z=X−μσ∼N(0,1).Z=\frac{X-\mu}{\sigma}\sim\mathrm N(0,1).

For a boundary xx, its standardised value is z=(x−μ)/σz=(x-\mu)/\sigma. Since σ>0\sigma>0, standardising does not reverse an inequality.

Not in the formula booklet: learn it.

The new variable ZZ is the standard Normal variable. It has no measurement units. Its mean is zero and its variance is one; every Normal model can therefore use the same probability table.

Worked example 1: a value above the mean

X∼N(80,42)X\sim\mathrm N(80,4^2). Standardise the boundary 8686, and rewrite P(X<86)\mathrm P(X<86) in terms of ZZ.

Plan. Subtract 8080 from the boundary and divide by 44.

z=86−804=1.5,P(X<86)=P(Z<1.5).z=\frac{86-80}{4}=1.5,\qquad \mathrm P(X<86)=\mathrm P(Z<1.5).

Answer. The boundary has z=1.5z=1.5; it is 1.51.5 standard deviations above the mean.

Check. Returning to the original scale gives 80+4(1.5)=8680+4(1.5)=86.

Change only the boundary to the same distance below the mean. Its standardised value must be negative.

Worked example 2: a value below the mean

X∼N(80,42)X\sim\mathrm N(80,4^2). Standardise 7474, and rewrite P(X<74)\mathrm P(X<74).

Plan. Use the signed difference between the boundary and the mean.

z=74−804=−1.5,P(X<74)=P(Z<−1.5).z=\frac{74-80}{4}=-1.5,\qquad \mathrm P(X<74)=\mathrm P(Z<-1.5).

Answer. The boundary has z=−1.5z=-1.5; it is 1.51.5 standard deviations below the mean.

Check. 80+4(−1.5)=7480+4(-1.5)=74. A negative standardised value correctly places it to the left of the mean.

Writing the standardisation also makes the choice of scale visible. Use the standard deviation in the denominator, even when the model is given with its variance.

Cumulative probabilities from tables

The supplied Normal distribution function table gives the area to the left of a standardised boundary. It does not give the area between zero and that boundary.

Cumulative Normal probability

For Z∼N(0,1)Z\sim\mathrm N(0,1),

Φ(z)=P(Z≤z).\Phi(z)=\mathrm P(Z\le z).

The booklet's table gives Φ(z)\Phi(z) for nonnegative zz. For example, the entry beside z=1.50z=1.50 is 0.93320.9332: that includes all the area to the left of zero as well as the area from zero to 1.501.50.

The table starts with Φ(0)=0.5000\Phi(0)=0.5000. To use a negative boundary, reflect its left tail into the equal right tail at the corresponding positive boundary:

Φ(−a)=1−Φ(a)(a≥0).\Phi(-a)=1-\Phi(a)\qquad(a\ge0).

The symmetry is about zero on the standardised scale.

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