probability in statistics

Question

The number of hours an American adult sleeps each night follows a Normal distribution with an average of 6.9 hours per night with a standard deviation of 0.68 hours. If we randomly select one American adult, what is the probability they sleep more than 5 hours per night?

Answer 1

A cursory look at the normal curve:

tells us the probability will be quite high. With a mean of 6.9, -1σ (1 standard deviation) would be 6.22 hrs and the probability would be 84.1%, -2σ would be 5.54 hrs and would be 97.7% and -3σ would be 4.86 hrs and would be 99.9%.

To get the exact number, using a TI-84 Plus CE calculator, use the "distr" button (2nd "vars"), then scroll down to "2: normalcdf("

For "lower" type "5" (this is the low end limit which the problem says is 5 hrs)

For "upper" type 1E99 (1, then the EE button (2nd comma), the 99) [this is the high limit - essentially infinity)

For "μ" type "6.9" (this is the mean)

For "σ" type "0.68" (this is the standard deviation)

Then scroll to "Paste" and hit enter twice. The answer is 0.9973978631 (in percent, that would be 99.74%)