0/1 points | previous answers devorestat9 4.e.041. ask your teacher my notes question part points submissions used 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 31 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.)
<span>0.3046% chance of damaging at least one of five payloads.
First, determine how many standard deviations it takes for the parachute to open at 100 meters. So
200 m - 100 m = 100 m
100 m / 31 m = 3.2258
Looking up on a standard normal table, the probability of having that deviation or higher is 0.00061, or 0.061%
Since you're looking for the probability of damaging at least one of five, the easiest way of calculating that is to subtract from 1 the probability of having all five parachutes successfully deploy. So
1 - (1 - 0.00061)^5 = 1 - 0.99939^5 = 1 - 0.996953719 = 0.003046281 = 0.3046%</span>
C. You'd be starting at a higher floor (positive number) and descending floors (negative number per x). If you assume x is time it's a little more logical and visual.
