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Loading Aircraft Before every flight, the pilot must verify that the total weight of the load is less than the maximum allowable load for the aircraft. The Bombardier Dash 8 aircraft can carry 37 passengers, and a flight has fuel and baggage that allows for a total passenger load of 6200 lb. The pilot sees that the plane is full and all passengers are men. The aircraft will be overloaded if the mean weight of the passengers is greater than \(6200 \mathrm{lb} / 37=167.6 \mathrm{lb}\). What is the probability that the aircraft is overloaded? Should the pilot take any action to correct for an overloaded aircraft? Assume that weights of men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb (based on Data Set I "Body Data" in Appendix B).

Step by step solution

01

Calculate the z-score

To find the probability that the mean weight of the passengers is greater than 167.6 lb, calculate the z-score using the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where:\( \bar{x} = 167.6 \, \mathrm{lb}\) (sample mean),\( \mu = 189 \, \mathrm{lb}\) (population mean),\( \sigma = 39 \, \mathrm{lb}\) (standard deviation),\( n = 37\) (sample size).Substitute the values into the formula.

02

Simplify the z-score calculation

Simplify the z-score calculation:\[ z = \frac{167.6 - 189}{\frac{39}{\sqrt{37}}} \]First calculate the standard error of the mean (SEM):\[ SEM = \frac{39}{\sqrt{37}} \]Then calculate the z-score.

03

Find the standard error

Calculate the standard error of the mean (SEM):\[ SEM = \frac{39}{\sqrt{37}} \approx 6.41 \]Now, calculate the z-score:\[ z = \frac{167.6 - 189}{6.41} \]

04

Compute the z-score

Plugging in the values, compute the z-score:\[ z = \frac{167.6 - 189}{6.41} \approx -3.34 \]

05

Determine the probability

Use the z-score table or a standard normal distribution calculator to find the probability corresponding to a z-score of -3.34. The probability is very close to 0 (almost zero).

06

Conclusion

Since the probability that the aircraft is overloaded is almost zero, the pilot does not need to take any action to correct for an overloaded aircraft.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Distribution

A probability distribution gives us a complete picture of how probabilities are distributed over different outcomes of a random variable. Essentially, it describes how often we can expect each different possible outcome if we were to repeat the random variable process many times. For instance, if the weights of men are normally distributed, we know how likely it is to randomly pick a man of a given weight.

For the pilot checking if the plane is overloaded, understanding the distribution of men’s weights helps in predicting whether the total passenger load exceeds the safe limit.

Probability distributions can take different forms, but a normal distribution, like in this case, tends to form a bell-shaped curve centered around the mean.

Normal Distribution

A normal distribution is a specific type of probability distribution that is symmetrical and bell-shaped. In the context of the original exercise, the weights of the men are normally distributed with a mean of 189 lb and a standard deviation of 39 lb.

Normal distributions are particularly useful in statistics because many real-world variables naturally follow this pattern. This helps in making predictions about a population from a sample. In our case, the mean weight of 37 men follows an expected pattern defined by the normal distribution.

When the weights of people or items follow a normal distribution, we can use z-scores to find probabilities, which are vital for determining if certain conditions (like overloading) are likely.

Standard Error

The standard error (SE) measures how much the sample mean weights are expected to fluctuate around the population mean weight. It's calculated using the formula: SE = \(\frac{\sigma}{\sqrt{n}}\) , where σ is the standard deviation and n is the sample size. For our problem, calculating SE helps us understand the variability in the average weight of 37 passengers from the known population mean weight.

In the example provided, the standard error is calculated as: \(SE = \frac{39}{\sqrt{37}} \approx 6.41\).

Having a smaller SE indicates that the sample mean is a more accurate reflection of the population mean, giving us confidence in the conclusions we draw from the data.

Sample Size

Sample size (n) is the number of observations in a sample. In our exercise, the sample size is 37 men, which means we are considering the weights of 37 individuals to calculate probabilities.

The sample size plays a crucial role in statistical analysis. Larger sample sizes generally provide more reliable estimates of population parameters because they reduce the margin of error and standard error.

In the z-score formula \(z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\), a larger sample size makes the denominator larger, which usually leads to a smaller standard error. This reduces the variation and can give more accurate z-scores, aiding in decision-making processes, like ensuring the plane isn’t overloaded.