Question

The respiratory disturbance index (RDI), a measure of sleep disturbance, for a specific population has a mean of 15 (sleep events per hour) and a standard deviation of 10. They are not normally distributed. Give your best estimate of the probability that a sample mean RDI of 100 people is between 14 and 16 events per hour?

Answer 1

Say \(X\) is the random variable for RDI, then you have to find
\[P(14<X<16)\]
Since the population isn't normally distributed, transforming to the standard normal distribution won't give an exact answer, so we use a slight variant for a good estimate.
\[Z=\frac{X-\mu}{\sigma}~~\Rightarrow~~Z\approx\frac{X-\mu}{\sigma/\sqrt n}\]
where \(n\) is the sample size.
\[P(14<X<16)\approx P\left(\frac{14-15}{10/\sqrt{100}}<\frac{X-15}{10/\sqrt{100}}<\frac{16-15}{10/\sqrt{100}}\right)\approx P(-1<Z<1)\]