Question

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 200 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 200 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? (Round your answer to four decimal places.)

Answer 1

Answer: 0.0082

Step-by-step explanation:

Let x be the random variable that represents the opening altitude.

We assume that opening altitude actually has a normal distribution .

As per given , we have

$\mu=200 \ \ \sigma=34$

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

z-score corresponds to x=100, $z=\dfrac{x-\mu}{\sigma}$

i.e.  $z=\dfrac{100-200}{34}\approx-2.94$

P-value = $P(z<100)=P(z<-2.94)=1-P(z<2.94)=1-0.9983589=0.0016411$

Let Y be a binomial variable that represents the opening altitude.

with parameters p=0.0016411 , n= 5

$P(x)=^nC_xp^x(1-p)^{n-x}$

Required probability :

$P(x\geq1)=1-P(x<1)\\\\=1-P(x=0)\\\\=1- ^5C_0(0.0016411)^0(1-0.0016411)^{5-0}\\\\=1-(1)(0.9983589)^5\\\\=1-0.99182138793\\\\=0.00817861207\approx0.0082$

Hence, the probability that there is equipment damage to the payload of at least one of five independently dropped parachutes = 0.0082