For beginner parachutists, the terminal velocity must be less than 5.00 m/s, and the parachutes used have a k-value that is normally distributed with a mean of 2.05/s and a standard deviation of 0.04/s. If the acceleration due to gravity is 9.81 m/s^2, determine the probability, to the nearest thousandth, that the terminal velocity is less than 5.00 m/s.
Please help me, the help in the last question was greatly appreciated. Thank-you!
Somebody please help me, I am really having problems. Thank-you!
Could you please post the relevant equations and your attempt at the solution?
What is a k-value for a parachute?
Thanks
Okay, relevant equations:
v(subscript)t=g/k
or
k=g/v(subscript)t
My attempt at the soloution, honestly, I have no clue, me and a friend have attempted this question and we still cannot figure it out. Please help me, I am totally lost.
so v mean = 9.81/2.05 = 4.785 m/s
k one sigma lower = 2.05 -0.04 = 2.01/s
velocity 1 sigma above mean = 9.81/2.01 = 4.881 /s
so
sigma of velocity = 4.881 - 4.785 = .095
now 5.00 is how many sigmas above mean?
5.00 - 4.785 = .215
and z = .215/.095 = 2.26 sigmas above mean (pretty improbable)
I only have a crude normal distribution table hgere but here are two values:
z = 2.2 then F(z) = .986
z = 2.3 then F(z) = .989
so we are talking around .988
You and your mate would be better off reading something about terminal velocity. You cannot post questions to Jiskha in the exam hall! :)
Here are some background reading:
this one from NASA
http://exploration.grc.nasa.gov/education/rocket/termvr.html
here's some lighter reading:
http://www.northallertoncoll.org.uk/physics/module%202/terminal%20velocity/terminal%20velocity.htm
and here's one on statistics:
http://en.wikipedia.org/wiki/Standard_deviation
It's the last one that concerns you most.
k=g/vt
so
vt=g/k
g=9.81 m/s/s
k=2.05 (mean value)
so the mean landing velocity is
vt=9.81/2.05=4.785
The minimum value (kmin) of k to have a landing velocity of 5.00 m/s is
5=9.81/km
or
kmin=9.81/5=1.962
Difference from mean
= 2.05-1.962
=0.088
Standard deviation (meausure of variability of the k-value)
= 0.04
Therefore, a parachute has to have a value of k at 0.088/0.04=2.2 standard deviations below the mean value to have a landing velocity higher than 5 m/s.
If you look up a table of normal distribution for a tail end of 2.2 standard deviations (sigma), you will find that the probability is 0.0139, i.e. there is a 1.39% chance that the skydivers will land at more than 5 m/s.
The table is available here:
http://www.math.unb.ca/~knight/utility/NormTble.htm
This is a quickie that does not really help you with your exams. Do your reading and prepare yourself accordings.
Good luck.
Thank-you, very much!!!
To determine the probability that the terminal velocity is less than 5.00 m/s, we need to calculate the z-score and convert it to a probability using the standard normal distribution table.
The formula to calculate the z-score is:
z = (x - μ) / σ
Where:
x = 5.00 m/s (terminal velocity)
μ = 2.05 m/s (mean of the k-value)
σ = 0.04 m/s (standard deviation of the k-value)
Substituting these values into the formula, we get:
z = (5.00 - 2.05) / 0.04
Now we can calculate the z-score:
z = 72.5
We need to find the probability associated with this z-score. However, it is not possible to directly find the probability for such a large z-score using the standard normal distribution table. Therefore, we can use an approximation by considering the z-score as significantly large.
Using the standard normal distribution table, we can find that the probability associated with a z-score of 3 is approximately 0.9987. Since our z-score is significantly larger than 3, we can approximate the probability as nearly 1.
Therefore, the probability that the terminal velocity is less than 5.00 m/s is approximately 1 (or 100%).
It is worth mentioning that while this approximation is valid for large z-scores, it may not be as accurate for smaller z-scores. For a more precise estimation, statistical software or tools can be used.