How often does the sample average overshoot?

Question

You take 25 independent measurements, each Normal with mean \mu and standard deviation \sigma. You compute their average. What is the probability that the sample average exceeds \mu + 0.5\sigma? Compare this to the chance a single measurement exceeds \mu + 0.5\sigma.

Accepted Answer

The average of n independent Normals is again Normal, with the same mean \mu but a shrunken standard deviation: \bar{X} \sim 	ext{Normal}\!\left(\mu,\ \frac{\sigma^2}{n}ight), \qquad 	ext{SD}(\bar{X}) = \frac{\sigma}{\sqrt{n}} = \frac{\sigma}{\sqrt{25}} = \frac{\sigma}{5}. Averaging pulls the spread in by a factor of \sqrt{n} = 5. We want P(\bar{X} \mu + 0.5\sigma). The z-score measures the overshoot in units of the average's standard deviation \sigma/5: z = \frac{0.5\sigma}{\sigma/5} = 0.5 	imes 5 = 2.5. From the normal table: P(\bar{X} \mu + 0.5\sigma) = P(Z 2.5) \approx 0.0062, about …

How often does the sample average overshoot?

Your answer

More Probability questions