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 m 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?

Step-by-step

Given:

mean, mu = 200 m

standard deviation, sigma = 30 m

sample size, N = 5

Maximum deviation for no damage, D = 100 m

Solution:

Z-score for maximum deviation

= (D-mu)/sigma

= (100-200)/30

= -10/3

From normal distribution tables, the probability of right tail with

Z= - 10/3

is 0.9995709, which represents the probability that the parachute will open at 100m or more.

Thus, by the multiplication rule, the probability that all five parachutes will ALL open at 100m or more is the product of the individual probabilities, i.e.

P(all five safe) = 0.9995709^5 = 0.9978565

So there is an approximately 1-0.9978565 = 0.214% probability that at least one of the five parachutes will open below 100m

answer:

it will cost 480 dollars to refurbish the larger wood floor

step-by-step explanation:

as it's a ratio of 4 that costs 384 dollars, divide the price by the ratio

384 / 4 = 96 dollars

96 dollars per unit of the ratio

now to get the price to refurbish the larger wood floor, times the price by the units

96 * 5 = 480

y=1/4x-4

step-by-step explanation: