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Airline passengers get heavier In response to the increasing weight of airline passengers, the Federal Aviation Administration (FAA) in $2003$told airlines to assume that passengers average $190$pounds in the summer, including clothes and carry-on baggage. But passengers vary, and the FAA did not specify a standard deviation. A reasonable standard deviation is $35$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 $30$passengers.

(a) Explain why you cannot calculate the probability that a randomly selected passenger weighs more than $200$pounds.

(b) Find the probability that the total weight of the passengers on a full flight exceeds $6000$pounds. Show your work. (Hint: To apply the central limit theorem, restate the problem in terms of the mean weight.)

Step by step solution

01

Part (a) Step 1: Given Information 

It is given in the question that, Population mean ($\mathrm{μ}$) $=190$

Population standard deviation $\left(\mathrm{σ}\right)=35$

sample size$\left(n\right)=30$

02

Part (a) Step 2: Explanation 

To calculate the probability the population must be normally distributed. Here, it has been supplied that the distribution of the population is not normally distributed. Therefore, the required probability cannot be computed here.

03

Part (b) Step 1: Given Information 

It is given in the question that, Population mean $\left(\mathrm{μ}\right)=190$

Population standard deviation $\left(\mathrm{σ}\right)=35$

sample size$\left(n\right)=30$

04

Part (b) Step 2: Explanation 

There are total passengers. The average weight of the passenger can be calculated as:

$\stackrel{-}{X}=\frac{6000}{30}$

$=200$

The probability that the total weight of the passengers in the flight is above pounds is calculated as follows:

$P(\stackrel{¯}{X}>200)=P(\frac{\stackrel{¯}{x}−\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{n}}}>\frac{200−\mathrm{μ}}{\frac{\mathrm{σ}}{\sqrt{n}}})$

$=P(Z>\frac{200−190}{\frac{35}{\sqrt{50}}})$

$=P(Z>1.56)\left(\text{From standard normal table}\right)$

$=0.0594$

Thus, the required probability is$0.0594$

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