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Example 2 referred to an elevator with a maximum capacity of \(4000 \mathrm{lb}\). When rating elevators, it is common to use a \(25 \%\) safety factor, so the elevator should actually be able to carry a load that is \(25 \%\) greater than the stated limit. The maximum capacity of \(4000 \mathrm{lb}\) becomes \(5000 \mathrm{lb}\) after it is increased by \(25 \%\), so 27 adult male passengers can have a mean weight of up to \(185 \mathrm{lb}\). If the elevator is loaded with 27 adult male passengers, find the probability that it is overloaded because they have a mean weight greater than \(185 \mathrm{lb}\). (As in Example 2, assume that weights of males are normally distributed with a mean of \(189 \mathrm{lb}\) and a standard deviation of 39 lb.) Does this elevator appear to be safe?

Step by step solution

01

Determine the mean weight needed

The problem states that for safety, the mean weight of the 27 passengers should be at most 185 lb. So, the mean weight limit per passenger is set at 185 lb.

02

State the given parameters

The weights of the male passengers are normally distributed with a mean \( \mu = 189 \text{lb} \) and a standard deviation \( \sigma = 39 \text{lb} \). The number of passengers \( n = 27 \).

03

Find the standard error of the mean

The standard error (SE) of the mean is given by the formula \[ SE = \frac{\sigma}{\sqrt{n}} \] Plugging in the values we get: \[ SE = \frac{39}{\sqrt{27}} = 7.507 \text{lb} \]

04

Calculate the z-score

The z-score is calculated using the formula \[ z = \frac{\text{X} - \mu}{SE} \] where \( \text{X} = 185 \text{lb}. \) Therefore, the z-score is \[ z = \frac{185 - 189}{7.507} = -0.533 \]

05

Determine the probability

Using the z-score table, find the probability corresponding to z = -0.533. This value is approximately 0.297.

06

Find the complement

The probability that the mean weight is greater than 185 lb is the complement of 0.297, which means \[ P(\overline{X} > 185) = 1 - 0.297 = 0.703 \]

07

Conclusion

Since the probability that the mean weight exceeds 185 lb is 70.3%, the elevator appears to be overloaded more often than not.

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Key Concepts

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

Normal Distribution

Normal distribution is fundamental for statistics, especially in safety engineering. It suggests that data tends to cluster around a central mean with symmetrical tails on either end. In our exercise, the weights of males are normally distributed with a mean of 189 lb and a standard deviation of 39 lb. This tells us most male weights are near the 189 lb mark, with fewer weights as we move away from the mean on either side. The bell curve illustrates normal distribution. The peak is the mean, and the width indicates the variability. Understanding this helps predict and calculate probabilities for various scenarios, like whether an elevator exceeds its weight limit.

Z-Score

The z-score helps by converting different data points into a common scale. It shows how far and in what direction a data point is from the mean, measured in standard deviations. In our example, the expected mean weight is 189 lb. However, we need to check if 27 passengers can have a mean weight of 185 lb and still be within safety limits. Using the formula \[ z = \frac{X - \mu}{SE} \], we calculate the z-score to see how unusual or typical this scenario is. We found a z-score of -0.533, indicating the mean weight of 185 lb is 0.533 standard deviations below the mean. A negative z-score shows the data point is below the mean.

Standard Error

Standard error (SE) is crucial in determining how precisely our sample mean represents the population mean. It is calculated by \[ SE = \frac{\sigma}{\sqrt{n}} \]. In our example, the calculation yields a SE of 7.507 lb. This error decreases as sample size (n) increases, meaning larger samples give a more accurate estimate of the population mean. Here, the SE helps gauge how much the average weight of 27 males deviates from the true population mean. With a mean weight of 185 lb for 27 passengers, we can now use the SE to calculate the z-score and evaluate probability.

Probability Calculation

Calculating probability is essential to determine safety in various scenarios. For our elevator problem, we need to find the probability that the average weight of 27 males exceeds 185 lb. After calculating the z-score as -0.533, we use the z-score table to find its corresponding probability. This is 0.297, meaning there's a 29.7% chance the mean weight is 185 lb or less. To find the probability of it being greater, we find the complement: \[ P(\overline{X} > 185) = 1 - 0.297 = 0.703 \]. Thus, there’s a 70.3% chance the elevator will be overloaded, making it unsafe.