The airline has found from past experience that the weight of checked-in baggage items is Normally distributed with mean 13 kg and standard deviation 3.1 kg.
Any assistance would be appreciated I have been working on it for ages and no luck. Thanks
Any assistance would be appreciated I have been working on it for ages and no luck. Thanks
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150kg for 10 means 15kg per baggage
standard error = 3.1 /sqrt 10 = 0.98
z -score = (15 - 13)/0.98 = 2.04
Pr(z>2.04) = 0.0207 to 4 dp
http://www.wolframalpha.com/input/?i=Pr%…
@antoine2
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it is the VARIANCE that would become 10 times, not the SD
standard error = 3.1 /sqrt 10 = 0.98
z -score = (15 - 13)/0.98 = 2.04
Pr(z>2.04) = 0.0207 to 4 dp
http://www.wolframalpha.com/input/?i=Pr%…
@antoine2
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it is the VARIANCE that would become 10 times, not the SD
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z score = (150 - 130)/31 = 0.645
prob = 1 - 0.74 = 0.26
prob = 1 - 0.74 = 0.26