AP Statistics Curriculum 2007 Bayesian Normal

Normal Example:

It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, with a standard deviation of 10 mph. Knowing that his car is capable of much higher speeds, the owner took the car to the shop. After a checkup, the speedometer was replaced with a new one, which gave a new estimate of 220 mph with a standard deviation of 4 mph. The errors are assumed to be normally distributed. We can say that the owner S’s prior beliefs about the top speed of his car were represented by:

µ ~ N(\(\mu_0\), \(\phi_0\)) = µ ~ N(185,\(10^2\))

We could then say that the measurements using the new speedometer result in a measurement of:

x ~ N(\(\mu\), \(\phi\)) = x ~ N(µ,\( 4^2\))

We note that the observation x turned out to be 210, and we see that S’s posterior beliefs about µ should be represented by:

µ | x ~ N(\(\mu_1\), \(\phi_1\))

where (rounded)

\(\phi_1\) = \((10^{-2} + 4^{-2})^{-1}\) = 14 = \(4^2\)

\(\mu_1\) = \(14(185/10^2 + 220/4^2) = 218\)

Therefore, the posterior for the top speed is:

\(\mu\) | x ~ N(\(218,4^2\))

Meaning 218 +/- 4 mph.

If the new speedometer measurements were considered by another person S’ who had no knowledge of the readings from the first speedometer, but still had a vague idea (from knowledge of the stock speedometer) that the top speed was about 200 +/- 30 mph, Then:

\(\mu\) ~ N(\(200,30^2\))

Then S’ would have a posterior variance:

\(\phi_1 = (30^{-2} + 4^{-2})^{-1} = 16 = 4^2\)

S’ would have a posterior mean of: