It is found that 99% of a particular design of sport parachute have k-values greater than or equal to 1.650/s. If the mean value of k is 1.730/s, determine, to the nearest thousandth, the standard deviation of the distribution of k-values.

Help, thank-you!

1 answer

It is basically the same problem as last night's, but in reverse. Your teacher wants to make sure you understand the subject.
Go back to the table of the Normal distribution:
http://www.math.unb.ca/~knight/utility/NormTble.htm
The table lists the probability of exceeding a particular value.
Z is the standard deviation (or sigma). Thus at Z=0, the probability is 0.5, meaning that 50% of the values are as likely to fall above or below the mean.
For Z=2.32 and 2.33, we have probabilities of 0.9898, 0.9901.
Thus a probability of 99% falls on Z=2.327.
With a mean value of 1.730, and 1.650 falls on the 99% mark, then there are 2.327 times the standard deviation between those limits.
Sigma (standard deviation)
=(1.730-1.650)/2.327
=0.034/s

I still think you would benefit by reading up on standard deviations, from your text-book, a reference book, or the third link I gave you last night:
http://en.wikipedia.org/wiki/Standard_deviation

In fact, after this problem, it would be beneficial to go back to yesterday's problem and you will find it quite easy.