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? Express your answer as a percentage point to the nearest percentage

Answer 1

The sample size is large enough (100) for the Central Limit Theorem to hold. Therefore the distribution of the sample mean is normal, and we can standardize the sample mean and obtain a standard normal random variable:
\[Z=\frac{\bar {X}-\mu}{\frac{\sigma}{\sqrt{n}}}\]
where Z ~ N(0, 1)
The required probability is found from:
\[P(14<\bar {X}<16)=P(\frac{14-15}{\frac{10}{\sqrt{100}}}<\frac{\bar {X}-15}{\frac{10}{\sqrt{100}}}<\frac{16-15}{\frac{10}{\sqrt{100}}})=P(-1<Z<1)\]
\[=0.3413+0.3413=0.6826\]
Expressed as a percentage we get 68.26%.