Question

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 185 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 185 and standard deviation 34 m. Equipment damage will occur if the parachute opens at an altitude of less than 100 m. What is the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes?

Answer 1

Solution:

   We are given that: The automatic opening device of a military cargo parachute has been designed to open when the parachute is 185 m above the ground.

Opening altitude actually has a Normal distribution with mean value 185 and standard deviation of 34 m.

That is: Mean =    and Standard Deviation =

Equipment damage will occur if the parachute opens at an altitude of less than 100 m.

We have to find the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

That is :

P( At least one damage out of five independently dropped parachutes ) = ........?

P( At least one damage out of five independently dropped parachutes ) = 1 - P( No One Damage )

where

P( No One Damage ) = P( All 5 are working )

P( No One Damage ) = P( all 5 parachute opens at an altitude of greater than 100 m )

P( No One Damage ) = [ P( X > 100 ) ]⁵

So first find : P( single parachute opens at an altitude of greater than 100 m) =...?

That is : P( X > 100 ) = .....?

Thus find z score for x = 100

Thus we get :

P( X > 100) = P( z > -2.50)

P( X > 100) = 1 - P( z < -2.50)

Look in z table for z = -2.5 and 0.00 and find corresponding area.

P( Z < -2.50) = 0.0062

Thus

P( X > 100) = 1 - P( z < -2.50)

P( X > 100) = 1 - 0.0062

P( X > 100) = 0.9938

Thus

P( No One Damage ) = [ P( X > 100 ) ]⁵

P( No One Damage ) = [ 0.9938 ) ]⁵

P( No One Damage ) = 0.9694

Thus

P( At least one damage out of five independently dropped parachutes ) = 1 - P( No One Damage )

P( At least one damage out of five independently dropped parachutes ) = 1 - 0.9694

P( At least one damage out of five independently dropped parachutes ) = 0.0306