The Normal distribution and inverse probabilities

What you’ll learn

• Standardise using the standard deviation.
• Calculate areas and inverse percentiles.
• Use symmetry and interpret Normal model assumptions.

Understand the idea

A Normal distribution is continuous, symmetric and centred on its mean. In $X\sim N(\mu,\sigma^2)$ the second parameter is the variance, so take its square root before standardising. The total area under the density is one, and a probability is an area rather than a curve height.

Subtract the mean and divide by the standard deviation to obtain a standard Normal variable. Draw or imagine the required interval before using a cumulative table or calculator. For an upper tail, subtract the cumulative probability from one; for an interval, subtract its two cumulative probabilities.

Inverse questions start from an area and find a boundary. Obtain the corresponding standard Normal quantile and then reverse the standardisation. If two percentiles are given, their two equations can determine an unknown mean and standard deviation. Check that the final boundary lies on the expected side of the mean.

Key formulae

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

$$P(a<X<b)=\Phi\!\left(\frac{b-\mu}{\sigma}\right)-\Phi\!\left(\frac{a-\mu}{\sigma}\right)$$

$$x_p=\mu+z_p\sigma,\qquad \Phi(z_p)=p$$

Worked example

A score is modelled by $X\sim N(70,8^2)$. Find the probability that a score exceeds 78.

1. The standard deviation is 8, so the boundary standardises to $(78-70)/8=1$.
2. The required region is the upper tail: $P(X>78)=P(Z>1)$.
3. Using $\Phi(1)=0.8413$, the probability is $1-0.8413=0.1587$.

Answer: $0.1587$ approximately.