SOLUTION: Please help me solve this problem: An Airline knows that the number of suitcases it loses per week on a certain route is a random variable having approximately the normal distribut
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Question 1176840: Please help me solve this problem: An Airline knows that the number of suitcases it loses per week on a certain route is a random variable having approximately the normal distribution with mean = 26.6 and SD = 5.8. Find the probability that in one week on the given route it will lose at least 22 suitcases. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! population mean = 26.5
population standard deviation = 5.8.
probability that a random variable taken from the population is greater than 22 is found by finding the z-score and then finding the proability of getting a z-score greater than that.
z-score = (x - m) / s
x is the randomly chosen score of 22.
m is the mean of 26.5
s is the standard deviation.
formula becomes z = (22 - 26.5) / 5.8 = -.775862069.
probability of getting a z-score higher than that is equal to the area under the normal distribution curve to the right of that z-score = .7810848541.
i used a calculator, but if you used a z-score table, you would have found:
z-score of -.77 has an area of .22065 to the left of it.
z-score of -.78 has an area of .21770 to the left of it.
through interpolation, you would have found that the area to the left of a z-score of -.775862069 would be equal to .2189206896.
the area to the right of that z-score would be 1 minus that = .7810793104.
that's very close to the area that the calculator gave as .7810848541.
obviously, the use of the calculator makes it much easier to get.
manual interpolation is much more labor intensive, much more subject to error if you're not extremely careful in your calculations, and can even be confusing when you're dealing with negative numbers as in this problem.
