Question

The automatic opening device of a military cargo parachute has been designed to open when the parachute is 155 m above the ground. Suppose opening altitude actually has a normal distribution with mean value 155 and standard deviation 30 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

the probability that one parachute of the five parachute is damaged is 0.156

Explanation:

From the given information;

Let consider X to be the altitude above the ground that a parachute opens

Then; we can posit that the probability that the parachute is damaged is:

P(X ≤ 100 )

Given that the population mean μ = 155

the standard deviation σ = 30

Then;

`P(X \leq 100 ) = ( (X- \mu)/(\sigma) \leq (100- \mu)/(\sigma))`

`P(X \leq 100 ) = ( (X- 155)/(30) \leq (100- 155)/(30))`

`P(X \leq 100 ) = (Z \leq (- 55)/(30))`

`P(X \leq 100 ) = (Z \leq -1.8333)`

`P(X \leq 100 ) = \Phi( -1.8333)`

From standard normal tables

`P(X \leq 100 ) = 0.0334`

Hence; the probability of the given parachute damaged is 0.0334

Let consider Q to be the dropped parachute

Given that the number of parachute be n= 5

The probability that the parachute opens in each trail be p = 0.0334

Now; the random variable Q follows the binomial distribution with parameters n= 5 and p = 0.0334

The probability mass function is:

Q
`\sim` B(5, 0.0334)

Similarly; the event that one parachute is damaged is :

Q ≥ 1

P( Q ≥ 1 ) = 1 - P( Q < 1 )

P( Q ≥ 1 ) = 1 - P( Y = 0 )

P( Q ≥ 1 ) = 1 - b(0;5; 0.0334 )

P( Q ≥ 1 ) =
`1 -(^5_0)* (0.0334)^0*(1-0.0334)^5`

P( Q ≥ 1 ) =
`1 -( (5!)/((5-0)!)) * (0.0334)^0*(1-0.0334)^5`

P( Q ≥ 1 ) = 1 - 0.8437891838

P( Q ≥ 1 ) = 0.1562108162

P( Q ≥ 1 )
`\approx` 0.156

Therefore; the probability that one parachute of the five parachute is damaged is 0.156