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Chapter 4: Problem 62

An electron emitter produces electron beams with changing kinetic energy that is uniformly distributed between three and seven joules. Suppose that it is possible to adjust the upper limit of the kinetic energy (currently set to seven joules). (a) What is the mean kinetic energy? (b) What is the variance of the kinetic energy? (c) What is the probability that an electron beam has a kinetic energy of exactly 3.2 joules? (d) What should be the upper limit so that the mean kinetic energy increases to eight joules? (e) What should be the upper limit so that the variance of kinetic energy decreases to 0.75 joules?

Short Answer

Expert verified
(a) 5 joules. (b) 1.3333 joules. (c) 0 probability. (d) 13 joules. (e) 6 joules.

Step by step solution

01

Mean Kinetic Energy Formula

For a uniform distribution between two values \(a\) and \(b\), the mean is calculated as \( \frac{a+b}{2} \). Here \(a = 3\) and \(b = 7\).
02

Calculate the Mean

Using the formula from Step 1, the mean kinetic energy is \( \frac{3+7}{2} = 5 \) joules.
03

Variance Formula for Uniform Distribution

The variance for a uniform distribution is calculated by the formula \( \frac{(b-a)^2}{12} \).
04

Calculate the Variance

Substituting \(a = 3\) and \(b = 7\) into the variance formula, we get \( \frac{(7-3)^2}{12} = \frac{16}{12} = \frac{4}{3} \approx 1.3333 \) joules.
05

Probability of Exact Value in Uniform Distribution

In a continuous uniform distribution, the probability of any single, exact value is zero.
06

Adjust Mean to 8

To increase the mean to 8 joules, set up the equation from mean formula \( \frac{3+b}{2} = 8 \).
07

Solve for New Upper Limit for Mean

Solving \( \frac{3+b}{2} = 8 \) gives \(3 + b = 16\) which means \(b = 13\).
08

Adjust Variance to 0.75

Set up the variance equation \( \frac{(b-a)^2}{12} = 0.75 \) where \(a = 3\).
09

Solve for New Upper Limit for Variance

Multiplying the variance equation by 12 gives \((b-3)^2 = 9\). Solving \(b-3 = 3\) or \(b-3 = -3\) results in \(b = 6\) or \(b = 0\). Only \(b = 6\) is feasible as an upper limit.

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Key Concepts

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

Mean Kinetic Energy
The mean kinetic energy is a measure of the average kinetic energy within a range of energies. For a uniform distribution, this range is defined by two limits: a lower limit, \(a\), and an upper limit, \(b\). The formula for calculating the mean of a uniform distribution is \( \frac{a+b}{2} \).
When we apply this formula to the given exercise where \(a = 3\) (lower limit) and \(b = 7\) (upper limit), the mean kinetic energy is calculated as:
\[ \frac{3+7}{2} = 5 \text{ joules} \]
This means that, on average, the electron beams produced have a kinetic energy of 5 joules.
Variance Calculation
Variance gives us an understanding of how much the kinetic energy values are spread out around the mean. In a statistical sense, it measures the "spread" of the data. For a uniform distribution between \(a\) and \(b\), the variance \(\sigma^2\) is calculated with the formula:
\[ \frac{(b-a)^2}{12} \]
By substituting the given limits where \(a = 3\) and \(b = 7\), we find:
\[ \frac{(7-3)^2}{12} = \frac{16}{12} = \frac{4}{3} \approx 1.3333 \text{ joules}^2 \]
This tells us that the spread of kinetic energy values from the mean is about 1.3333 joules squared.
Probability of Exact Value
In continuous uniform distributions, the probability of obtaining any exact value is always zero. This is a distinct characteristic of continuous distributions compared to discrete ones. This is because continuous distributions encompass an infinite number of possible outcomes within any range.
For example, even though the emitter produces energies between 3 and 7 joules, the probability of obtaining exactly 3.2 joules as the kinetic energy is \(0\). There are infinitely many possibilities between each joule. Thus, the precise selection of any single point becomes negligibly small, resulting in a probability of zero.
Adjusting Distribution Parameters
Sometimes, it becomes necessary to adjust distribution parameters, like the upper limit, to modify the characteristics of the distribution. Such adjustments may be necessary to achieve a specific mean or variance.
To increase the mean kinetic energy to 8 joules, we need to change \(b\) using the mean formula \( \frac{3 + b}{2} = 8 \). Solving this gives:
  • \(3 + b = 16\)
  • \(b = 13\)
Therefore, setting the upper limit to 13 increases the mean to the desired 8 joules.
On the other hand, to decrease the variance to 0.75, we apply the formula \( \frac{(b-a)^2}{12} = 0.75 \). Solving for \(b\) provides:
  • \((b-3)^2 = 9\)
  • Thus, \(b-3 = 3\) or \(b-3 = -3\)
The feasible solution is \(b = 6\). Adjusting to this upper limit changes the variance as required.

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