Question

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 175 m above the ground. suppose opening altitude actually has a normal distribution with mean value 175 and standard deviation 35 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? (give your answer to four decimal places.)

Answer 1

Combination of normal and binomial distributions.

Calculate probability of opening below 100m.
mean=175 m
standard deviation = 35 m
z=(100-175)/35=-15/7
p=P(Z<z)=0.01606229   from normal probability tables.
Calculate probability of at least one parachute opens below 100m
p=0.01606229n=5
x=0
P(X>0)
=1-P(x=0)
=1-C(5,0)p^x(1-p)^5
=1-1*0.01606229^0*(1-0.01606229)^5
=1-0.9222 (approximately)
=0.0778

Answer:
Probability that at least one of five cargoes will be damaged is 0.0778