*Ask Uncle Colin* is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to colin@flyingcoloursmaths.co.uk and Uncle Colin will do what he can.

Dear Uncle Colin,

My research has determined that female adults have a mean overhead reach of 208.5cm, with a standard deviation of 8.6cm, and follows a normal distribution.

I wanted to know the probability that the mean overhead reach of 50 female adults would lie between 180cm and 200cm and got a crazy small answer of about $2 \times 10^{-13}$. What am I doing wrong?

-- Got Adults Under Stretching, Somehow

Dear GAUSS,

That looks about right, actually! You would expect the mean of a sample of 50 stretchers to follow a normal distribution with a mean of 208.5cm and a standard deviation of $\frac{8.6}{\sqrt{50}}\text{cm} \approx 1.22$cm.

The z-score for a mean of 180cm is -23.4, and the z-score for 200cm is about -6.99. You would turn to your normal distribution tables and say "oh pooh. Those numbers aren't on my tables." Instead, you turn to Wolfram|Alpha or (better) Abramovitz and Stegun, and find that the associated probabilities are $9.8 \times 10^{-122}$ and $1.4 \times 10^{-12}$. The difference between those is, well, about $1.4 \times 10^{-12}$ -- the first number is minuscule in comparison.

Does this number look about right? Well, to have an observed mean of 50 observations a full standard deviation below the expected mean is (very rough ballpark) about the same as the probability of all 50 being below the mean. That would be a probability of $2^{-50} \approx 10^{-15}$, which is a similar magnitude.

Hope that helps!

-- Uncle Colin

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008.
He lives with an espresso pot and nothing to prove.