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.

Question

Help, thank-you!

Answer 1

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.