Standard Deviation, Statistics??

Question

A resort tracks the number of days each guest stays. They've discovered that this variable is normally distributed with a mean of 9 days and a standard deviation of 3. They would expect 10% of the guests stay longer than how many days?

Answer 1

There is a very simple way to do this.
Tables of the standard normal variable Z show that z > 1.282 has a probability of 10%.
Convert to your normal distribution using the formula
z = (x - mean) / (standard deviation)
backwards to find an x value.

Answer 2

You need to use the cumulative distribution function. 10% of guests stay longer than x days, where x has a cumulative distribution of 90%. For a normal distribution, cdf = 0.5*(1 + erf((x - μ) / (σ*sqrt(2)))). μ is the mean, 9, and σ is the standard deviation, 3. erf is the error function, which is tabulated.

You know that you want cdf = 0.9, which lets you solve algebraically for erf((x - μ) / (σ*sqrt(2))). Then you can look up what argument you need for erf, and you can solve for (x - μ) / (σ*sqrt(2)) equal to that argument.

There are two other ways to do this. You may be able to find a tabulation of confidence intervals for normal distributions, which are essentially a special case of the above. If we want to know how many days have 10% of guests staying longer, it's the number of standard deviations that has a confidence interval of 80% (because it also leaves out 10% on the bottom side). Multiply that number by 3. (This is the approach described by the answerer below me.)

The final way is really more of an estimate. (1 − 1/k^2)*100% of values should be within k standard deviations of the mean. So for a confidence interval of 80%, we need (1 − 1/k^2)*100 = 80 ==> 1 - 1/k^2 = 0.8 ==> 1/k^2 = 0.2 ==> k^2 = 5 ==> k = sqrt(5). If we're looking for sqrt(5) standard deviations above the mean, that's 3*sqrt(5), and therefore the number of days that is the lower bound of 10% of guest stays will be 9 + 3*sqrt(5) = 15.7 days, or about 16 days. As I said, though, this is only an estimate.