Answer:
the probability that one parachute of the five parachute is damaged is 0.156
Step-by-step 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;
[tex]P(X \leq 100 ) = ( \dfrac{X- \mu}{\sigma} \leq \dfrac{100- \mu}{\sigma})[/tex]
[tex]P(X \leq 100 ) = ( \dfrac{X- 155}{30} \leq \dfrac{100- 155}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq \dfrac{- 55}{30})[/tex]
[tex]P(X \leq 100 ) = (Z \leq -1.8333)[/tex]
[tex]P(X \leq 100 ) = \Phi( -1.8333)[/tex]
From standard normal tables
[tex]P(X \leq 100 ) = 0.0334[/tex]
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 [tex]\sim[/tex] 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 ) = [tex]1 -(^5_0)* (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = [tex]1 -( \dfrac{5!}{(5-0)!}) * (0.0334)^0*(1-0.0334)^5[/tex]
P( Q ≥ 1 ) = 1 - 0.8437891838
P( Q ≥ 1 ) = 0.1562108162
P( Q ≥ 1 ) [tex]\approx[/tex] 0.156
Therefore; the probability that one parachute of the five parachute is damaged is 0.156
