Answer :
In order to design an elevator that can safely carry 15 people, assuming a worst-case scenario of 15 male passengers, we need to calculate the maximum total allowable weight with a 0.98 probability that this maximum will not be exceeded.
To find the maximum total allowable weight, we can use the properties of the normal distribution and z-scores. Given that the mean weight of males is 179 lb and the standard deviation is 33 lb, we can calculate the z-score corresponding to a probability of 0.98.
Using a standard normal distribution table or a calculator, we find that the z-score for a probability of 0.98 is approximately 2.05. The z-score represents the number of standard deviations away from the mean.
To calculate the maximum total allowable weight, we multiply the z-score by the standard deviation and add it to the mean weight:
maximum weight = mean + (z-score * standard deviation)
maximum weight = 179 + (2.05 * 33)
maximum weight ≈ 179 + 67.65
maximum weight ≈ 246.65
Therefore, the maximum total allowable weight, rounded to the nearest whole number, is approximately 247 lb. This means that in order to have a 0.98 probability that the weight limit will not be exceeded when 15 male passengers are randomly selected, the maximum weight the elevator should safely carry is 247 lb.
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Final answer:
To find the maximum total weight for an elevator designed to safely carry 15 men, we use the central limit theorem and normally distributed weight data. The maximum weight is calculated using the formula X = μ*15 + z*σ*sqrt(15) where X is the maximum weight, μ is the mean weight, z is the z-score corresponding to the desired probability of 0.98, and σ is the standard deviation. This value is then rounded to the nearest whole number for practical use.
Explanation:
In this problem, we can make use of the Central Limit Theorem which states that the sum of a large number of independent and identically distributed measurements will have approximately a normal distribution, regardless of the shape of the original measurement distribution.
Given that the weight distribution of male staff members is normally distributed with a mean (μ) of 179 lb and a standard deviation (σ) of 33 lb, we're looking for the maximum total allowable weight for 15 male members which correlates to a 0.98 probability. A worst-case scenario would involve the heaviest 15 men, so we would use the statistics for the men.
We want to find an X such that P(sum of weights ≤ X)=0.98. The sum of weights of 15 males can be modeled as a normal distribution with mean = 15μ and standard deviation= sqrt(15)σ. We will convert X to a z-score by subtracting the mean and dividing by the standard deviation which we will compare with z=2.05(corresponding to a 0.98 probability).
Therefore X= μ*15 + z*σ*sqrt(15) gives us the desired maximum weight when rounded to the nearest whole number.
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