In an elevator, the maximum occupancy and maximum weight allowed are listed as 26 passengers and 4500 lbs, respectively. If the distribution of the weight of an individual is an uniform random variable between 110 lbs and 220 lbs, calculate the probability that the total weight of the passengers exceeding the maximum weight allowed under maximum occupancy? (Assume that if n ≥ 25, the Central Limit Theorem can be used)
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I will need to assume that the passengers' weights are independent to solve this problem.
A uniform distribution on [a, b] has mean (a + b)/2 and standard deviation
(b - a)/sqrt(12).
The mean weight (in lbs) of a given passenger is (110 + 220)/2 = 165.
The standard deviation of the weight of a given passenger (in lbs) is
(220 - 110)/sqrt(12) = 110/sqrt(12).
The mean of the total weight (in lbs) of 26 passengers is (165)(26) = 4290.
Since I'm assuming the weights are independent, the standard deviation of the total weight (in lbs) of 26 passengers is (110/sqrt(12)) * sqrt(26) = 110sqrt(13/6).
Because 26 > 25, the Central Limit Theorem implies that the total weight (in lbs) of 26 passengers is approximately normally distributed with mean 4290 and standard deviation 110sqrt(13/6).
Let Z represent the standard normal random variable with mean 0 and standard deviation 1.
The probability that the total weight of 26 passengers exceeds 4500 lbs is about
P(Z > (4500 - 4290)/(110sqrt(13/6)))
= 1 - P(Z <= (4500 - 4290)/(110sqrt(13/6)))
= 1 - P(Z <= 1.30)
= 1 - 0.9032 from the normal table
= 0.0968
Lord bless you today!
A uniform distribution on [a, b] has mean (a + b)/2 and standard deviation
(b - a)/sqrt(12).
The mean weight (in lbs) of a given passenger is (110 + 220)/2 = 165.
The standard deviation of the weight of a given passenger (in lbs) is
(220 - 110)/sqrt(12) = 110/sqrt(12).
The mean of the total weight (in lbs) of 26 passengers is (165)(26) = 4290.
Since I'm assuming the weights are independent, the standard deviation of the total weight (in lbs) of 26 passengers is (110/sqrt(12)) * sqrt(26) = 110sqrt(13/6).
Because 26 > 25, the Central Limit Theorem implies that the total weight (in lbs) of 26 passengers is approximately normally distributed with mean 4290 and standard deviation 110sqrt(13/6).
Let Z represent the standard normal random variable with mean 0 and standard deviation 1.
The probability that the total weight of 26 passengers exceeds 4500 lbs is about
P(Z > (4500 - 4290)/(110sqrt(13/6)))
= 1 - P(Z <= (4500 - 4290)/(110sqrt(13/6)))
= 1 - P(Z <= 1.30)
= 1 - 0.9032 from the normal table
= 0.0968
Lord bless you today!