Answer:
the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215
Step-by-step explanation:
Let consider Q to be the opening altitude.
The mean μ = 135 m
The standard deviation = 35 m
The probability that the equipment damage will occur if the parachute opens at an altitude of less than 100 m can be computed as follows:
[tex]P(Q<100) = P(\dfrac{X- 135}{\sigma} < \dfrac{100 - 135}{35}})[/tex]
[tex]P(Q<100) = P(z< \dfrac{-35}{35}})[/tex]
[tex]P(Q<100) = P(z<-1)[/tex]
[tex]P(Q<100) = 0.1587[/tex]
If we represent R to be the number of parachutes which have equipment damage to the payload out of 5 parachutes dropped.
The probability of success = 0.1587
the number of independent parachute n = 5
the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes can be computed as:
P(R ≥ 1) = 1 - P(R < 1)
P(R ≥ 1) = 1 - P(R = 0)
The probability mass function of the binomial expression is:
P(R ≥ 1) = [tex]1 - (^5_0)(0.1587)^0(1-0.1587)^{5-0}[/tex]
P(R ≥ 1) =[tex]1 - (\dfrac{5!}{(5-0)!})(0.1587)^0(1-0.1587)^{5-0}[/tex]
P(R ≥ 1) = 1 - 0.5785
P(R ≥ 1) = 0.4215
Hence, the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes is 0.4215
