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StClowers
 one year ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb. and 211 lb. The new populations of pilots has normally distributed weights with a mean of 150 lbs. and a standard deviation of 34.4 lbs.
a. If a pilot is randomly selected, find the probability that his weight is between 150 lb. and 211 lb. The probability is approximately?
StClowers
 one year ago
An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 150 lb. and 211 lb. The new populations of pilots has normally distributed weights with a mean of 150 lbs. and a standard deviation of 34.4 lbs. a. If a pilot is randomly selected, find the probability that his weight is between 150 lb. and 211 lb. The probability is approximately?

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whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Okay, if the mean is 150 lbs and the std dev is 34.4 pounds, 68.2% of the pilots will have a weight within 1 std dev, or in the range 15034.4= 115.6 to 150+34.4 = 184.4 95.45% will have a weight within 2 std deviation, or the range 1502*34.4 = 81.2 to 150+2*34.4=218.8 With a normal distribution, the population is equally balanced on each side of the mean, so we can take half of those percentages to get the section between 150 (the mean) and 211. Therefore, about 47% or so should fall within 150 and 211, or P = 0.47. Some of the lighter pilots may want to put on lead underwear :)

StClowers
 one year ago
Best ResponseYou've already chosen the best response.0I put this answer in and it said it was wrong :/

StClowers
 one year ago
Best ResponseYou've already chosen the best response.0its asking for the central limit theorem

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0Hmm. Well, I wonder how exact they want it to be? 95.45/2 = 47.725, but that goes out to slightly beyond 211 pounds. Central Limit Theorem is what allows us to use the 68.2/95.45/97.3 rule of thumb about the distribution around the mean. In any case, I'm sorry that you got that marked wrong, and am curious what the correct answer is supposed to be.

StClowers
 one year ago
Best ResponseYou've already chosen the best response.0on the previous ones ive had to do, it is asking for the cumulative area from the left

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0okay, but here we are only interested in the area from the mean, to the mean+2 sigma.

StClowers
 one year ago
Best ResponseYou've already chosen the best response.0ok so it gave me the right answer of .5224 but not sure how they did that

whpalmer4
 one year ago
Best ResponseYou've already chosen the best response.0well, 10.5224 = 0.4776 or 47.76%. I don't see how they are getting > 0.5, however!
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