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I think I can! An important measure of the performance of a locomotive is its "adhesion," which is the locomotive's pulling force as a multiple of its weight. The adhesion of one 4400 -horsepower diesel locomotive varies in actual use according to a Normal distribution with mean \(\mu=0.37\) and standard deviation \(\sigma=0.04\) (a) For a certain small train's daily route, the locomotive needs to have an adhesion of at least 0.30 for the train to arrive at its destination on time. On what proportion of days will this happen? Show your method. (b) An adhesion greater than 0.50 for the locomotive will result in a problem because the train will arrive too early at a switch point along the route. On what proportion of days will this happen? Show your method. (c) Compare your answers to parts (a) and (b). Does it make sense to try to make one of these values larger than the other? Why or why not?

Short Answer

Expert verified
(a) 95.99%, (b) 0.06%. It doesn't make sense to prefer one higher due to different contexts.

Step by step solution

01

Understanding the Problem

We're dealing with a Normal distribution of adhesion for a locomotive where the mean adhesion is \( \mu = 0.37 \) and the standard deviation is \( \sigma = 0.04 \). We need to calculate the probabilities of adhesion being at least 0.30 and more than 0.50.
02

Finding Probability for Part (a)

We want to find the probability that the adhesion is at least 0.30. We convert the adhesion level to a Z-score using the formula \( Z = \frac{X - \mu}{\sigma} \). For \( X = 0.30 \), the Z-score \( Z = \frac{0.30 - 0.37}{0.04} = -1.75 \). Using the standard normal distribution table, find the probability for \( Z = -1.75 \). This gives us \( P(Z < -1.75) \). Since we need \( P(Z \geq -1.75) \), we use \( 1 - P(Z < -1.75) \).
03

Calculating Probability for Part (a)

Using the standard normal distribution table, \( P(Z < -1.75) \approx 0.0401 \). Therefore, \( P(Z \geq -1.75) = 1 - 0.0401 = 0.9599 \). Thus, the proportion of days the adhesion is at least 0.30 is approximately 0.9599 or 95.99%.
04

Finding Probability for Part (b)

We need to find the probability that the adhesion is greater than 0.50. We convert this to a Z-score: \( Z = \frac{0.50 - 0.37}{0.04} = 3.25 \). Using the standard normal distribution table, we find the probability for \( Z = 3.25 \). We are looking for \( P(Z > 3.25) \).
05

Calculating Probability for Part (b)

From the standard normal distribution table, \( P(Z < 3.25) \approx 0.9994 \). Thus, \( P(Z > 3.25) = 1 - 0.9994 = 0.0006 \). Hence, the proportion of days the adhesion is greater than 0.50 is approximately 0.0006 or 0.06%.
06

Comparing Results for Part (c)

In part (a), 95.99% of days have adhesion at least 0.30, while in part (b), only 0.06% of days have adhesion greater than 0.50. It does not make sense to strive for one value to be larger than the other, as part (a) ensures the locomotive performs effectively, while part (b) is about preventing a rare issue.

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Key Concepts

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

Z-score
A Z-score is a statistical measurement that describes a data point's relation to the mean of a group of points. It shows how many standard deviations an element is from the mean. In the context of our locomotive's adhesion example, a Z-score helps us understand how far a specific adhesion level is from the average adhesion level of 0.37.

To calculate a Z-score, use the formula:
  • division{X - \mu}{\sigma}
where:
  • \(X\) is the adhesion level we want the Z-score for,
  • \(\mu\) is the mean adhesion (0.37),
  • \(\sigma\) is the standard deviation of adhesion (0.04).
The Z-score provides a way to determine how likely a particular adhesion level is within the normal distribution by comparing it to the overall average and variability.
Probability Calculation
Once we have calculated the Z-score, the next step is to use it to determine the probability of observing this adhesion level or a more extreme one. This is where probability calculation comes in. We use the standard normal distribution table, often known as the Z-table, to find probabilities associated with different Z-scores.

For adhesion levels like 0.30 and 0.50 in our locomotive example, we calculate the Z-scores and then use these values to look up probabilities in the Z-table. Here's what the steps look like:
  • Calculate the Z-score for the adhesion level you're interested in.
  • Use the Z-score to find the area under the curve to the left of this Z-score, which represents the probability \(P(Z < Z\text{-score})\).
  • For probabilities greater than a certain level, subtract: \( P(Z > Z\text{-score}) = 1 - P(Z < Z\text{-score}) \).
In real-world scenarios, this gives us insight into the likelihood of different adhesion levels based on the normal distribution.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution with a mean of 0 and a standard deviation of 1. It allows us to easily look up probabilities and Z-scores using tables, making calculations straightforward once an individual Z-score is determined.

In practice, the raw data is first converted using a Z-score formula into a statistic that fits the standard normal distribution model. This transformation achieves a uniform scale of measurement for different datasets. Each point tells us how many standard deviations an observation is from the mean.

Using the standard normal distribution helps us predict how data behaves under a normal curve. It simplifies the analysis of diverse data sets, which is why it's so commonly applied in probability and statistics. Understanding this concept is key, especially in problems like the locomotive's adhesion, where reliable predictions affect planning and performance management.

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