S1.6 - The Normal distribution

Syllabus
2019
Topic
S1.6
Level
AS

Use the Normal distribution

A Normal distribution is a continuous, bell-shaped model that is symmetric about its mean μ\mu. Write XN(μ,σ2)X\sim N(\mu,\sigma^2), where σ2\sigma^2 is the variance and σ>0\sigma>0 is the standard deviation. Symmetry makes the mean, median and mode coincide, and probabilities are areas under the curve.

Z=Xμσ,ZN(0,1)Z=\frac{X-\mu}{\sigma},\qquad Z\sim N(0,1)

Required event Standard-Normal calculation
P(X<a)P(X<a) Φ ⁣(aμσ)\Phi\!\left(\dfrac{a-\mu}{\sigma}\right)
P(X>a)P(X>a) 1Φ ⁣(aμσ)1-\Phi\!\left(\dfrac{a-\mu}{\sigma}\right)
P(a<X<b)P(a<X<b) Φ ⁣(bμσ)Φ ⁣(aμσ)\Phi\!\left(\dfrac{b-\mu}{\sigma}\right)-\Phi\!\left(\dfrac{a-\mu}{\sigma}\right)
equal tails about μ\mu use symmetry: Φ(z)=1Φ(z)\Phi(-z)=1-\Phi(z)

Here Φ(z)=P(Zz)\Phi(z)=P(Z\le z) is read from the cumulative Normal table or calculator. For XN(210,252)X\sim N(210,25^2),P(190<X<240)=Φ(1.2)Φ(0.8)=0.88490.2119=0.6730.P(190<X<240)=\Phi(1.2)-\Phi(-0.8)=0.8849-0.2119=0.6730.Draw the event mentally before using the table: a right tail needs a complement, while an interval needs a difference.

For an unknown boundary or parameter, turn the stated cumulative probability into its zz-value and use x=μ+zσx=\mu+z\sigma. If P(X<152)=0.05P(X<152)=0.05 and P(X<180)=0.60P(X<180)=0.60, then152=μ1.6449σ,180=μ+0.2533σ.152=\mu-1.6449\sigma,\qquad180=\mu+0.2533\sigma.Subtracting gives σ=28/1.898214.8\sigma=28/1.8982\approx14.8, then μ176\mu\approx176. The same method finds percentiles, quartiles and symmetric limits.

After finding a single-observation probability pp, ordinary probability rules still apply. For independent observations, use products or a complement such as 1(1p)n1-(1-p)^n. For a conditional event, divide the joint probability by the probability of the condition; the Normal table supplies the component areas, not the final conditional ratio.

Check that a symmetric bell-shaped model is plausible; marked skewness undermines a Normal model. Because the distribution is continuous, P(X=a)=0P(X=a)=0, so strict and inclusive endpoints give the same probability. S1.6 requires shape, symmetry and cumulative-table use, but not the density formula, derivations or interpolation between table entries.