anonymous
  • anonymous
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?
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

whpalmer4
  • whpalmer4
Okay, 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 150-34.4= 115.6 to 150+34.4 = 184.4 95.45% will have a weight within 2 std deviation, or the range 150-2*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 :-)
anonymous
  • anonymous
I put this answer in and it said it was wrong :-/
anonymous
  • anonymous
its asking for the central limit theorem

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

whpalmer4
  • whpalmer4
Hmm. 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.
anonymous
  • anonymous
on the previous ones ive had to do, it is asking for the cumulative area from the left
whpalmer4
  • whpalmer4
okay, but here we are only interested in the area from the mean, to the mean+2 sigma.
anonymous
  • anonymous
ok so it gave me the right answer of .5224 but not sure how they did that
whpalmer4
  • whpalmer4
well, 1-0.5224 = 0.4776 or 47.76%. I don't see how they are getting > 0.5, however!
anonymous
  • anonymous
me either

Looking for something else?

Not the answer you are looking for? Search for more explanations.