Probability wind speed exceeds 13.5


New Member
I am sooo stuck on this question...
Suppose the average wind speed equals 12 miles per hour, and that wind speeds are normally distributed. the standard deviation is 3.4 miles per hour. Find the probability that the wind speed will exceed 13.5 miles per hour.
( I am using a TI-83plus)
This is what I am thinking...
Normalcdf(.10, 13.5, 12, 3.4) = .67 or 67%
I have already been told this is not right! I am not looking for the correct answer on a silver platter. Just looking for directions in solving this. Pleeez!
I think you're interpreting the calculator's results incorrectly, although you're on the right track.

I don't know the details of calculator commands, but it seems you're using the Cumulative Distribution Function (CDF) command, so what you're getting from the calculator is the percentage of wind speeds 13.5 mph and under (assuming that the this is in fact what the calculator is doing).

So this is a simple fix. If you want to find the percentage of wind speeds above 13.5, subtract .67 from 1:

1 - .67 = .33

Is this correct?

Hope this helps.