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Chapter 8: Problem 21

An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage checked by an individual passenger is a random variable \(x\) with a mean value of \(50 \mathrm{lb}\) and a standard deviation of \(20 \mathrm{lb}\). If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With \(n=100\), the total weight exceeds the limit when the average weight \(\bar{x}\) exceeds \(6000 / 100\).)

Short Answer

Expert verified
The approximate probability that the total weight of the passengers' baggage will exceed the limit is virtually 0.

Step by step solution

01

Calculate the average weight limit per passenger

This problem refers to the average weight limit per passenger, which is obtained by dividing the total baggage limit by the total number of passengers. Let's denote this as \(\bar{x}_{limit}\). Hence, \(\bar{x}_{limit} = \frac{6000}{100} = 60\) lbs.
02

Calculate the mean and the standard deviation of the total weight

According to the Central Limit Theorem, the sum or average of a large number of independent and identically distributed random variables will have a distribution that is approximately normal. Here, each passenger's baggage weight can be seen as one such random variable. Therefore, the mean and the standard deviation of the passengers' total weight, when normalized to the number of passengers, would also be the mean and the standard deviation of this variable. That's \(\mu = 50 lbs\) and \(\sigma = \frac{20 lbs}{\sqrt{100}} = 2 lbs\) respectively.
03

Convert to a Z-Score

The Z-score is a way of standardizing a random variable. It counts the number of standard deviations a data point (in this case, \(\bar{x}_{limit}\)) is away from the mean. For our case, the Z-score will be \( Z = \frac{{\bar{x}_{limit} - \mu}}{{\sigma}} = \frac{{60 lbs - 50 lbs}}{{2 lbs}} = 5 \)
04

Calculate the probability

The probability that the total weight of the baggage will exceed the limit is the same as the probability that Z is more than 5. However, for standard normal tables, it reaches a maximum around Z = 3.4. Beyond this, the area under the curve virtually becomes 1. Hence, we can say that the probability is virtually 0 that the weight will exceed the limit.

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Most popular questions from this chapter

A random sample is selected from a population with mean \(\mu=100\) and standard deviation \(\sigma=10 .\) Determine the mean and standard deviation of the \(\bar{x}\) sampling distribution for each of the following sample sizes: a. \(n=9\) b. \(n=15\) c. \(n=36\) d. \(n=50\) c. \(n=100\) f. \(n=400\)

Explain the difference between a population characteristic and a statistic.

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