In response to the increasing weight of airline passengers, the Federal Aviation

Question

In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 189 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 34 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 21 passengers. What is the approximate probability that the total weight of the passengers exceeds 4369 pounds? (Round your answer to four decimal places.) In response to the increasing weight of airline passengers, the Federal Aviation Administration in 2003 told airlines to assume that passengers average 189 pounds in the summer, including clothing and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is 34 pounds. Weights are not Normally distributed, especially when the population includes both men and women, but they are not very non-Normal. A commuter plane carries 21 passengers. What is the approximate probability that the total weight of the passengers exceeds 4369 pounds? (Round your answer to four decimal places.)

Explanation / Answer

Here population mean=189 and sd= 34 Using Central limit theorem , we can find x bar= 4369/21=208.05 and sample sd= 34/sqrt(21)=7.419

The test statistic, z= (x bar - mean)/sample sd = (208.05-189)/7.419=2.57

We Need to find P(x bar > 208.05)=?

So, P(x bar > 208.05)=P(z>2.57)=1 - P(z<2.57) = 1-0.99492 = 0.0051

The required probability is 0.0051

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