91Ó°ÊÓ

In einem Schmelzofen sollen Gold und Kupfer getrennt werden. Dazu muss der Ofen auf jeden Fall eine Temperatur von weniger als \(1083^{\circ} \mathrm{C}\) haben, da dies der Schmelzpunkt von Kupfer ist. Der Schmelzpunkt von Gold liegt bei \(1064{ }^{\circ} \mathrm{C}\). Um die Temperatur im Schmelzofen zu bestimmen, wird eine Messsonde benutzt. Ist \(\mu\) die tatsächliche Temperatur im Schmelzofen, so sind die Messwerte \(X\) der Sonde normalverteilt mit Erwartungswert \(\mu\) und Varianz \(\sigma^{2}=25\). Der Schmelzofen ist betriebsbereit, wenn die Temperatur \(\mu\) über dem Schmelzpunkt von Gold aber noch unter den Schmelzpunkt des Kupfers liegt. Die Entscheidung, ob der Ofen betriebsbereit ist, wird mithilfe der Messsonde bestimmt. Dabei wird so vorgegangen, dass der Ofen als betriebsbereit erklärt und mit dem Einschmelzen begonnen wird, wenn die Messsonde einen Messwert zwischen 1064 und \(1070{ }^{\circ} \mathrm{C}\) anzeigt. (a) Wie groß ist die Wahrscheinlichkeit, dass bei diesem Vorgehen der Ofen irrtümlich für betriebsbereit erklärt wird, wenn die Temperatur mindestens \(1083^{\circ} \mathrm{C}\) beträgt? (b) Wie groß ist die Wahrscheinlichkeit, dass die Temperatur im Ofen bei diesem Vorgehen den Schmelzpunkt des Goldes nicht überschreitet? (c) Ist es möglich eine Wahrscheinlichkeit dafür anzugeben, dass die Temperatur im Hochofen zwischen 1064 und \(1083^{\circ} \mathrm{C}\) liegt?

Step by step solution

01

Recall the z-score formula

The z-score formula is given as: \(z=\frac{X-\mu}{\sigma}\), where \(\mu\) is the mean value and \(\sigma\) is the standard deviation. For our problem, the variance \(\sigma^2 = 25\), so the standard deviation \(\sigma = 5\).

02

Calculate the z-scores

For each question, we have to calculate the z-score to find the corresponding probability. (a) We want to find the probability that the probe reads a temperature between \(1064{ }^{\circ} \mathrm{C}\) and \(1070{ }^{\circ} \mathrm{C}\), given that the actual temperature is at least \(1083^{\circ} \mathrm{C}\). We'll start by finding the z-scores of \(1064\) and \(1070\). For \(X = 1064\), the z-score is: $$z_{1064} = \frac{1064 - 1083}{5} = \frac{-19}{5}=-3.8$$ For \(X = 1070\), the z-score is: $$z_{1070} = \frac{1070 - 1083}{5} = \frac{-13}{5}=-2.6$$ (b) To find the probability that the gold's melting point is not reached, we need to find the z-score for \(X = 1064\). As computed in (a), \(z_{1064} = -3.8\).

03

Find the probabilities

Using the z-scores we found in Step 2, we will now find the probabilities. (a) Using the z-table (or a calculator), we find that \(P(Z \le -3.8) \approx 0.0001\) and \(P(Z \le -2.6) \approx 0.0047\). The probability that the probe reads a temperature between \(1064{ }^{\circ} \mathrm{C}\) and \(1070{ }^{\circ} \mathrm{C}\) when the actual temperature is at least \(1083^{\circ} \mathrm{C}\) is the difference between these two probabilities: \(P(-3.8 \le Z \le -2.6) \approx 0.0047 - 0.0001 = 0.0046\) (b) Using the z-table (or a calculator), we find that \(P(Z \le -3.8) \approx 0.0001\). That is, there is about a \(0.0001\) probability that the gold's melting point is not reached when the actual temperature is not at least \(1083^{\circ} \mathrm{C}\). (c) It is not possible to give a direct probability for the temperature in the furnace to be between \(1064{ }^{\circ} \mathrm{C}\) and \(1083^{\circ} \mathrm{C}\), since the normal distribution is based on a continuous function and not on discrete values. The probability could be interpreted as the area under the normal distribution curve, but this would require knowing additional information about the distribution of temperatures when the furnace is in operation.

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Key Concepts

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

Understanding Z-Score

The z-score is a statistical measure that tells us how many standard deviations an element is from the mean. In the context of our exercise, the z-score helps us understand where a certain temperature measurement lands in relation to the overall temperature distribution within the furnace.

In simple terms, a z-score indicates whether a measurement is typical (closer to zero), unusually low (negative z-score), or unusually high (positive z-score). To calculate a z-score, we use the formula \(z=\frac{X-\mu}{\sigma}\), where \(X\) is the value we're examining, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For the given problem, the variance \(\sigma^2 = 25\), so the standard deviation \(\sigma = 5\).

Understanding how to compute and interpret z-scores is crucial for answering questions about the furnace's operation and ensuring the separation of gold and copper at the right temperatures.

Probability in Context

Probability is a way of expressing the likelihood of an event happening. It is a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In our exercise, the probability concept is essential to determine the chances that the furnace temperature is within a certain range.

When we speak about the probability that the furnace is incorrectly declared ready, we're essentially asking, 'What is the likelihood that the temperature reading falls between 1064°C and 1070°C when the actual temperature is at least 1083°C?' This is measured using the area between the corresponding z-scores on a normal distribution curve.

By finding the probability associated with specific z-scores obtained from temperature readings, we can make informed decisions about when to start the smelting process.

The Importance of Melting Point

The melting point of a substance is the temperature at which it changes from a solid to a liquid state. It is a crucial physical property, particularly in smelting processes like the one described in our exercise.

In our scenario, the melting points of gold and copper dictate the necessary conditions for the furnace operation. Gold melts at 1064°C, and copper at 1083°C. This information helps establish the temperature range within which the furnace should ideally operate to separate the two metals efficiently and safely.

Understanding the melting point is vital to avoid melting copper when we only want to melt gold. Controlling the furnace temperature to stay between these melting points ensures a successful separation process.

Variance — Gauging Temperature Variability

Variance is a statistical measurement of the spread between numbers in a data set. More formally, it measures how far each number in the set is from the mean and thus from every other number in the set. In the case of our furnace, the variance \(\sigma^2 = 25\) gives us an idea of how temperature measurements by the probe might fluctuate around the actual furnace temperature \(\mu\).

A smaller variance would mean temperature readings by the probe are more consistently close to the actual temperature, whereas a larger variance would indicate more significant fluctuations. In practical terms, knowing the variance helps in assessing how reliable the temperature readings are. This is crucial for making accurate decisions when dealing with processes that have narrow temperature requirements, such as the smelting of gold and copper.