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Suppose that \(Y\) possesses the density function $$f(y)=\left\\{\begin{array}{ll}c y, & 0 \leq y \leq 2 ,\\\0, & \text { elsewhere .}\end{array}\right.$$ a. Find the value of \(c\) that makes \(f(y)\) a probability density function. b. Find \(F(y)\). c. Graph \(f(y)\) and \(F(y)\). d. Use \(F(y)\) to find \(P(1 \leq Y \leq 2)\). e. Use \(f(y)\) and geometry to find \(P(1 \leq Y \leq 2)\).

Short Answer

Expert verified
c is \(\frac{1}{2}\). \(F(y) = \begin{cases} 0, & y < 0 \\ \frac{1}{4}y^2, & 0 \leq y \leq 2 \\ 1, & y > 2 \end{cases}\). \(P(1 \leq Y \leq 2) = \frac{3}{4}\).

Step by step solution

01

Finding the value of c

To make \(f(y)\) a probability density function, we need to normalize it such that the total probability integral over the support equals 1:\[\int_{0}^{2} c y \, dy = 1\]Computing the integral:\[\left[ \frac{c y^2}{2} \right]_{0}^{2} = 1\]\[\frac{c (2^2)}{2} - \frac{c (0^2)}{2} = 1\]\[2c = 1\]Thus, solving for \(c\), we get \(c = \frac{1}{2}\).
02

Finding the cumulative distribution function (CDF), F(y)

The cumulative distribution function \(F(y)\) is found by integrating the density function from 0 to \(y\):\[F(y) = \int_{0}^{y} \frac{1}{2} t \, dt\]Computing the integral:\[F(y) = \left[ \frac{1}{4} t^2 \right]_{0}^{y} = \frac{1}{4} y^2\]Thus, the CDF is:\[F(y) = \begin{cases} 0, & y < 0 \\frac{1}{4}y^2, & 0 \leq y \leq 2 \1, & y > 2\end{cases}\]
03

Graphing f(y) and F(y)

For \(f(y)\), plot the function \(f(y) = \frac{1}{2}y\) for \(0 \leq y \leq 2\), where it is a line increasing linearly from the origin to the point (2,1). Beyond this interval, \(f(y) = 0\). For \(F(y)\), plot \(F(y) = \frac{1}{4}y^2\) for \(0 \leq y \leq 2\). It is a parabolic curve starting at the origin and increasing to 1 at \(y = 2\). Beyond this interval, \(F(y)\) is 0 for \(y < 0\) and 1 for \(y > 2\).
04

Using F(y) to find P(1 ≤ Y ≤ 2)

The probability \(P(1 \leq Y \leq 2)\) can be found using:\[P(1 \leq Y \leq 2) = F(2) - F(1)\]Using the CDF equations:\[ F(2) = 1 \F(1) = \frac{1}{4} \\]\[P(1 \leq Y \leq 2) = 1 - \frac{1}{4} = \frac{3}{4}\]
05

Using f(y) and geometry to find P(1 ≤ Y ≤ 2)

The probability \(P(1 \leq Y \leq 2)\) can also be found using the area under \(f(y)\) between 1 and 2:\[P(1 \leq Y \leq 2) = \int_{1}^{2} \frac{1}{2} y \, dy\]Computing the integral:\[= \left[ \frac{1}{4} y^2 \right]_{1}^{2} \]\[= \frac{1}{4} (2^2) - \frac{1}{4} (1^2) \]\[= \frac{1}{4} \times 4 - \frac{1}{4} \times 1\]\[= 1 - \frac{1}{4} = \frac{3}{4}\]

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Key Concepts

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

Cumulative Distribution Function
The cumulative distribution function (CDF), denoted as \( F(y) \), is a key concept in probability that provides the probability that a random variable \( Y \) is less than or equal to a specific value \( y \). In simpler terms, it gives us the cumulative probability up to a point. To find the CDF from a probability density function (PDF), we integrate the PDF over the range from the lowest value up to \( y \). This integration process accumulates the probabilities into the CDF. For our specific function, integrating \( f(y) \) from 0 to \( y \) results in \( F(y) = \frac{1}{4}y^2 \) for \( 0 \leq y \leq 2 \). This squared relationship implies that the CDF isn't a straight line, but rather a curve that steepens as \( y \) increases, reflecting increasing total probability. The CDF is 0 for \( y < 0 \) and 1 for \( y > 2 \), as the probability reaches its full value in the defined range.
Integration in Probability
Integration in probability is a fundamental tool, especially when working with continuous random variables. The idea here is to sum up probabilities over an interval to find total probabilities or to derive other related functions like the CDF. In our problem, we integrated the PDF \( f(y) = \frac{1}{2}y \) from 0 to \( y \) to find the CDF, \( F(y) = \frac{1}{4}y^2 \). This means we're accumulating the probability density into a total probability up to \( y \). When dealing with PDFs, remember that the entire area under the curve over its defined range must equal 1 to satisfy the properties of a probability density function. This explains why the integral of a PDF over its entire support gives 1. Thus, mastering basic integration techniques helps us solve various problems involving continuous random variables.
Graphing Probability Functions
Graphing probability functions provides a visual understanding of how probabilities are distributed. For a probability density function like \( f(y) = \frac{1}{2}y \), the graph shows a line starting from the origin and rising linearly to the point (2,1). Beyond the range \( 0 \leq y \leq 2 \), the function is zero, meaning no probability density. On the other hand, the cumulative distribution function \( F(y) = \frac{1}{4}y^2 \) forms a parabolic curve from the origin to the point (2,1), showing how probabilities accumulate to reach 1 as \( y \) increases to 2. Graphing these functions helps in identifying key characteristics like increasing trends and can highlight visual differences, like linear versus quadratic growth. Remember, for the PDF, the area under the curve signifies probability, while for the CDF, the curve itself directly indicates total probability at each \( y \).
Calculating Probabilities
Calculating probabilities using both the PDF and CDF provides flexibility and deeper understanding. For example, to find \( P(1 \leq Y \leq 2) \), we can use the CDF: \( P(1 \leq Y \leq 2) = F(2) - F(1) \). Substituting the values from the CDF, \( F(2) = 1 \) and \( F(1) = \frac{1}{4} \), yields \( P(1 \leq Y \leq 2) = \frac{3}{4} \). This calculation shows the proportion of the distribution lying between these bounds. Alternatively, using integration and geometry, this probability can also be obtained by calculating the area under \( f(y) \) from 1 to 2: \( \int_{1}^{2} \frac{1}{2}y \, dy \). This approach effectively measures the weighted area between these points, reinforcing the concept of probability in the context of area under a curve. Both methods should give the same result if calculated correctly, demonstrating consistency and choice in problem-solving.

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Most popular questions from this chapter

When we discussed the \(\chi^{2}\) distribution in this section, we presented (with justification to follow in Chapter 6 ) the fact that if \(Y\) is gamma distributed with \(\alpha=n / 2\) for some integer \(n\), then \(2 Y / \beta\) has a \(\chi^{2}\) distribution. In particular, it was stated that when \(\alpha=1.5\) and \(\beta=4\) \(W=Y / 2\) has a \(\chi^{2}\) distribution with 3 degrees of freedom. a. Use the applet Gamma Probabilities and Quantiles to find \(P(Y<3.5)\). b. Use the applet Gamma Probabilities and Quantiles to find \(P(W<1.75) .\) [Hint: Recall that the \(x^{2}\) distribution with \(\nu \text { degrees of freedom is just a gamma distribution with } \alpha=\nu / 2 \text { and } \beta=2 .]\) c. Compare your answers to parts (a) and (b).

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The length of time \(Y\) necessary to complete a key operation in the construction of houses has an exponential distribution with mean 10 hours. The formula \(C=100+40 Y+3 Y^{2}\) relates the cost \(C\) of completing this operation to the square of the time to completion. Find the mean and variance of \(C\).

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