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The current in a certain circuit as measured by an ammeter is a continuous random variable \(X\) with the following density function: $$ f(x)=\left\\{\begin{array}{cl} .075 x+.2 & 3 \leq x \leq 5 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Graph the pdf and verify that the total area under the density curve is indeed 1 . b. Calculate \(P(X \leq 4)\). How does this probability compare to \(P(X<4)\) ? c. Calculate \(P(3.5 \leq X \leq 4.5)\) and also \(P(4.5

Short Answer

Expert verified
The total area is 1. Both \(P(X \leq 4)\) and \(P(X < 4)\) are 0.4625. \(P(3.5 \leq X \leq 4.5)\) and \(P(4.5 < X)\) are both 0.5.

Step by step solution

01

Verify Total Area Under the Curve

To verify that the given function is a probability density function, we need to show that the total area under the curve is 1. The given density function is:\[f(x) = \begin{cases} 0.075x + 0.2 & \text{for } 3 \leq x \leq 5, \0 & \text{otherwise.}\end{cases}\]Integrate \( f(x) \) from 3 to 5:\[\int_3^5 (0.075x + 0.2) \, dx = \left[\frac{0.075}{2}x^2 + 0.2x\right]_3^5.\]Calculating this integral gives:\[\left(\frac{0.075}{2} (5)^2 + 0.2(5)\right) - \left(\frac{0.075}{2} (3)^2 + 0.2(3)\right)= (0.9375 + 1) - (0.3375 + 0.6) = 1 - 0.9375 = 1.\]Hence, the area under the curve is indeed 1.
02

Graph the PDF

To graph the probability density function, plot the linear function \(0.075x + 0.2\) between \(x = 3\) and \(x = 5\). The function starts at \(0.075 \times 3 + 0.2 = 0.425\) at \(x = 3\) and ends at \(0.075 \times 5 + 0.2 = 0.575\) at \(x = 5\). The line rises from left to right between these points.
03

Calculate \(P(X \leq 4)\)

To find \(P(X \leq 4)\), integrate \(f(x)\) from 3 to 4:\[P(X \leq 4) = \int_3^4 (0.075x + 0.2) \, dx = \left[ \frac{0.075}{2}x^2 + 0.2x \right]_3^4.\]Evaluating the integral:\[\left( \frac{0.075}{2} \times 16 + 0.8 \right) - \left( \frac{0.075}{2} \times 9 + 0.6 \right) = (0.6 + 0.8) - (0.3375 + 0.6) = 1.4 - 0.9375 = 0.4625.\]
04

Compare \(P(X \leq 4)\) and \(P(X < 4)\)

Since \(X\) is a continuous random variable, \(P(X \leq 4) = P(X < 4)\) due to the property of continuous distributions that probability at a single point is zero. Thus, \(P(X < 4) = 0.4625\).
05

Calculate \(P(3.5 \leq X \leq 4.5)\)

Integrate \(f(x)\) from 3.5 to 4.5:\[P(3.5 \leq X \leq 4.5) = \int_{3.5}^{4.5} (0.075x + 0.2) \, dx = \left[ \frac{0.075}{2}x^2 + 0.2x \right]_{3.5}^{4.5}.\]Calculating this:\[\left( \frac{0.075}{2} \times (4.5)^2 + 0.9 \right) - \left( \frac{0.075}{2} \times (3.5)^2 + 0.7 \right) = (0.759375 + 0.9) - (0.459375 + 0.7) = 1.659375 - 1.159375 = 0.5.\]
06

Calculate \(P(4.5 < X)\)

To find \(P(4.5 < X)\), integrate \(f(x)\) from 4.5 to 5:\[P(4.5 < X) = \int_{4.5}^{5} (0.075x + 0.2) \, dx = \left[ \frac{0.075}{2}x^2 + 0.2x \right]_{4.5}^{5}.\]Evaluating this:\[\left( \frac{0.075}{2} \times (5)^2 + 1 \right) - \left( \frac{0.075}{2} \times (4.5)^2 + 0.9 \right) = (1.46875) - (0.759375 + 0.9) = 0.5.\]

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Key Concepts

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

Continuous Random Variables
A continuous random variable is a type of random variable that can take on an infinite number of possible values within a given range. In contrast to discrete random variables, which have countable outcomes, continuous random variables are associated with measurements like length, temperature, or, in this case, electric current. These variables are depicted through a probability density function (PDF), which helps in determining the probability of the variable falling within a specific interval.
The key characteristic of continuous random variables is that the probability of the variable taking on any specific, single value is zero. This is because there are infinitely many possible points within any given range. So, when we look at probabilities for continuous random variables, we are interested in the probability across an interval, such as between two values. This is where integration plays a crucial role to calculate these probabilities over intervals.
Integration in Probability
Integration is an essential tool in probability when dealing with continuous random variables. Since these variables have a range of values, we use integration to calculate the total probability over an interval. This process involves integrating the probability density function (PDF) over a specific range of values. The result provides the cumulative probability that the random variable falls within a certain interval.
For a continuous random variable with a PDF of \( f(x) \), the probability that \( X \) lies between two points, \( a \) and \( b \), is given by the integral \( \int_{a}^{b} f(x) \, dx \). The integral of the PDF over its entire domain must be equal to 1, which represents 100% certainty that the variable takes on some value within its possible range. Understanding how to perform these integrations allows us to handle questions like computing the probability \( P(X \leq 4) \), or any other similar probability.
Properties of Continuous Distributions
The properties of continuous distributions are important in understanding how they differ from discrete ones. A major property is that with continuous variables, the probability of a random variable exactly equaling a specific value is always zero. Instead, we calculate probabilities over intervals.
One property is that the total area under the PDF curve for a continuous random variable must equal 1. This ensures that the random variable accounts for all possible outcomes in its range, much like how the sum of probabilities for discrete outcomes sums up to 1.
  • Another key property is that the PDF is non-negative. This makes sense because probabilities cannot be negative.
  • Also, the differences between cumulative probabilities, \( P(X \leq a) \) and \( P(X < a) \), are nonexistent for continuous distributions, unlike in the discrete case. This is because the probability at any single point, like \( P(X = a) \), is zero.
These properties make continuous distributions a really flexible and powerful tool in statistics and probability, particularly when dealing with measurements or scenarios that do not have a finite set of outcomes.

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Most popular questions from this chapter

Suppose Appendix Table A.3 contained \(\Phi(z)\) only for \(z \geq 0\). Explain how you could still compute a. \(P(-1.72 \leq Z \leq-.55)\) b. \(P(-1.72 \leq Z \leq .55)\) Is it necessary to tabulate \(\Phi(z)\) for \(z\) negative? What property of the standard normal curve justifies your answer?

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In response to concerns about nutritional contents of fast foods, McDonald's has announced that it will use a new cooking oil for its french fries that will decrease substantially trans fatty acid levels and increase the amount of more beneficial polyunsaturated fat. The company claims that 97 out of 100 people cannot detect a difference in taste between the new and old oils. Assuming that this figure is correct (as a long-run proportion), what is the approximate probability that in a random sample of 1000 individuals who have purchased fries at McDonald's, a. At least 40 can taste the difference between the two oils? b. At most \(5 \%\) can taste the difference between the two oils?

The article "Reliability of Domestic-Waste Biofilm Reactors" (J. of Envir. Engr., 1995: 785-790) suggests that substrate concentration \(\left(\mathrm{mg} / \mathrm{cm}^{3}\right)\) of influent to a reactor is normally distributed with \(\mu=.30\) and \(\sigma=.06\). a. What is the probability that the concentration exceeds \(.50 ?\) b. What is the probability that the concentration is at most \(.20\) ? c. How would you characterize the largest \(5 \%\) of all concentration values?

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