2.04e–2.04g - Normal models, standardisation and inverse probabilities

2.04e–2.04g - Normal models, standardisation and inverse probabilities

Use the mean and variance to interpret a normal model (2.04e), calculate probabilities and inverse boundaries (2.04f), and connect its curve with histograms, spread and familiar proportions (2.04g). The aim is to turn a measurement question into the right area, rather than trust an unexplained calculator output.

The mean centres the curve and the standard deviation sets its spread

A machine cuts rods with mean length 50 mm and standard deviation 4 mm. Under a normal model, what proportion exceed 56 mm? Six millimetres sounds small, but its significance depends on the spread: it is one and a half standard deviations above the mean.

You already know

Variance is the square of standard deviation. In a density histogram, area represents frequency; after scaling the total area to 1, it represents observed proportions. A model describes chance behaviour rather than guaranteeing a sample's exact frequencies.

The curve packages an entire distribution of measurements into a centre and spread.

Normal distribution

A normal distribution is a continuous, symmetric, bell-shaped probability model. In X∼N(μ,σ2)X\sim\mathrm N(\mu,\sigma^2), μ\mu is the mean and σ2\sigma^2 the variance, with σ>0\sigma>0. Its symmetry line is x=μx=\mu; its inflection points, where the curve changes between convex and concave, lie at x=μ±σx=\mu\pm\sigma.

For lengths XX in millimetres, the rod model is X∼N(50,16)X\sim\mathrm N(50,16). Its standard deviation is 4 mm, not 16 mm. The inflections are at 46 and 54 mm. The total area is 1; area between bounds gives their probability, whereas the curve's height is a density, not a probability.

Normal rod-length model with inflections and the one-, two- and three-standard-deviation intervals

A surprisingly large majority fits within two standard deviations, but a normal model has no finite cut-off at three: the remaining tail probability is small, not zero.

Normal proportions and inflections

For a normal distribution,

intervalapproximate proportionμ−σ<X<μ+σ68.3%μ−2σ<X<μ+2σ95.4%μ−3σ<X<μ+3σ99.7%\begin{array}{c|c} \text{interval}&\text{approximate proportion}\\ \mu-\sigma<X<\mu+\sigma&68.3\%\\ \mu-2\sigma<X<\mu+2\sigma&95.4\%\\ \mu-3\sigma<X<\mu+3\sigma&99.7\% \end{array}

Thus about two-thirds lie within one standard deviation, about 95% within two and almost all within three. Inflections occur at x=μ±σx=\mu\pm\sigma. These facts are not printed as such on the supplied sheet. Not in the formula sheet: learn it.

The corresponding rod-length intervals are 46–54, 42–58 and 38–62 mm. These proportions belong to the normal model; a different distribution need not follow them. Since an exact single value has zero probability in a continuous model, P(X<54)=P(X≤54)\mathrm P(X<54)=\mathrm P(X\leq54). This contrasts with the discrete spikes in the previous lesson.

Histograms from increasingly large samples of a normal population tend to reveal the normal shape more clearly, with suitable bins and random variation still present. Scale histogram area to 1 for comparison with the probability curve. Increasing σ\sigma widens the curve and lowers its peak, keeping total area 1; changing μ\mu shifts it horizontally. You do not need the equation of the normal curve. A symmetric sample alone does not establish every tail prediction, and a normal model can assign tiny probabilities to physically impossible negative measurements; assess whether that matters in context.

Key idea

The mean locates the curve; the standard deviation sets distances along the measurement axis. Probability comes from area, never from reading a height.

Use the second parameter correctly before deciding which interval is appropriate.

Standardisation measures distance in standard deviations

Subtract the centre and divide by the spread to express a measurement's position on one common scale.

Standardising a normal variable

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

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

This is the standard normal distribution. Single-observation standardisation is not printed on the sheet; the sheet's expression for a sample mean uses a different standard deviation. Not in the formula sheet: learn it.

The result is unit-free, and the positive denominator preserves inequalities. Write Φ(z)=P(Z≤z)\Phi(z)=\mathrm P(Z\leq z). Symmetry gives Φ(−z)=1−Φ(z)\Phi(-z)=1-\Phi(z); in particular, Φ(0)=1/2\Phi(0)=1/2. A calculator can also evaluate areas directly from the stated mean and standard deviation.

Worked example 1: a lower-tail area

Question. For rod lengths X∼N(50,16)X\sim\mathrm N(50,16), find P(X<56)\mathrm P(X<56) to three significant figures.

Plan. Standardise 56, then evaluate the area below that value.

z=56−504=1.5use the standard deviationP(X<56)=Φ(1.5)lower-tail area=0.9331927…normal calculator function\begin{aligned} z&=\frac{56-50}{4}=1.5 && \text{use the standard deviation} \\ \mathrm P(X<56)&=\Phi(1.5) && \text{lower-tail area} \\ &=0.9331927\ldots && \text{normal calculator function} \end{aligned}

Why this step? Φ(1.5)\Phi(1.5) is already the area below the required cut-off; no complement is needed.

Answer. The probability is 0.933, to three significant figures.

Check. The upper tail is 1−0.9331927…=0.0668072…1-0.9331927\ldots=0.0668072\ldots, answering the opening question. It is smaller than the roughly 16% above one standard deviation.

Change only the lower bound of the region: keep 56 as the upper cut-off but now start at 46 instead of extending indefinitely to the left.

Continue this lesson.

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