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Suppose that \(Y\) has density function f(y)=\left\\{\begin{array}{ll}k y(1-y), & 0 \leq y \leq 1, \\\0, & \text { elsewhere }\end{array}\right. a. Find the value of \(k\) that makes \(f(y)\) a probability density function. b. Find \(P(.4 \leq Y \leq 1)\). c. Find \(P(.4 \leq Y<1)\). d. Find \(P(Y \leq .4 | Y \leq .8)\) e. Find \(P(Y<.4 | Y<.8)\)

Short Answer

Expert verified
a) \(k=6\); b) 0.92; c) 0.92; d) 0.24; e) 0.24.

Step by step solution

01

Define Condition for PDF

To determine the constant \(k\) that makes \(f(y)\) a probability density function (pdf), the integral of \(f(y)\) over its entire range must equal 1. Therefore, we solve the equation: \[ \int_0^1 ky(1-y) \; dy = 1. \]
02

Solve Integral for k

Compute the integral: \( \int_0^1 ky(1-y) \; dy \) by expanding the expression inside: \( ky(1-y) = ky - ky^2 \). We integrate term by term: \[ \int_0^1 ky - ky^2 \; dy = k \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^1 = k \left( \frac{1}{2} - \frac{1}{3} \right) = k \left( \frac{1}{6} \right). \] Therefore, \( k \times \frac{1}{6} = 1 \) which implies \( k = 6 \).
03

Calculate P(.4 ≤ Y ≤ 1)

With \(k = 6\), find \(P(.4 \leq Y \leq 1)\) by computing the integral \( \int_{0.4}^1 6y(1-y) \; dy \). Expand and integrate: \[ \int_{0.4}^1 6y - 6y^2 \; dy = 6 \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_{0.4}^1. \] Calculate: \[ 6 \left[ \left( \frac{1}{2} - \frac{1}{3} \right) - \left( \frac{0.16}{2} - \frac{0.064}{3} \right) \right] = 6 \left[ \frac{1}{6} - \frac{0.08 - 0.02133}{6} \right]. \] Continue simplifying to get \(6 \times 0.153333 = 0.92\).
04

Calculate P(.4 ≤ Y < 1)

Since the probability of a continuous distribution taking a specific value is zero, \( P(.4 \leq Y < 1) = P(.4 \leq Y \leq 1) = 0.92 \).
05

Calculate P(Y ≤ .4 | Y ≤ .8)

Compute the conditional probability using \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), where \( A = \{Y \leq 0.4\} \) and \( B = \{Y \leq 0.8\} \). First, find \(P(Y ≤ 0.4)\) and \(P(Y ≤ 0.8)\):\[ P(Y \leq 0.4) = \int_0^{0.4} 6y(1-y) \, dy = 6 \left[ \frac{y^2}{2} - \frac{y^3}{3} \right]_0^{0.4} = 0.153333. \]\[ P(Y \leq 0.8) = \int_0^{0.8} 6y(1-y) \, dy = 6 \left[ 0.256 \right] = 1.536. \] Then \(P(Y ≤ .4 | Y ≤ .8) = \frac{0.153333}{0.64} \approx 0.24\).
06

Calculate P(Y < .4 | Y < .8)

For continuous distributions, \( P(Y < .4 | Y < .8) = P(Y \leq .4 | Y \leq .8) = 0.24 \) since the inclusion or exclusion of the exact endpoint does not affect probability.

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Key Concepts

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

conditional probability
Conditional probability is a way of finding the probability of an event (let's call it Event A) given that another event (Event B) has already occurred. It's a fundamental concept in probability theory that helps us understand how probabilities can change based on new information. For example, to calculate the conditional probability of Event A given Event B, we use the formula: \( P(A|B) = \frac{P(A \cap B)}{P(B)} \). This means we divide the probability of both events happening, \( P(A \cap B) \), by the probability of Event B, \( P(B) \).

In our scenario, this concept comes into play when we want to find the probability of \( Y \leq 0.4 \) given that \( Y \leq 0.8 \). We first compute the probabilities individually, such as \( P(Y \leq 0.4) \) and \( P(Y \leq 0.8) \), using integration. Then, by applying the formula above, it helps us "adjust" the probability of the first event, given the certain situation of the second event.

It's a powerful tool for understanding how probabilities can shift when one condition or scenario changes to another, offering a deeper insight into the interconnectedness of events.
integration
Integration is a mathematical process used to find areas under curves. In statistics, particularly within the context of probability density functions (pdfs), integration aids us in determining probabilities across continuous distributions. When dealing with pdfs, the total area under the curve (usually over its entire range) should equal 1, representing certainty of all possible outcomes.

In practice, the integration helps compute probabilities for certain ranges of a distribution. For instance, if you want to calculate the probability of the variable taking values between \( 0.4 \) and \( 1 \), you need to integrate the density function over this range. The resulting area is the probability.
  • To find the entire probability, compute: \( \int_0^1 ky(1-y) \, dy = 1 \), ensuring the function's total area is 1.
  • For a specific range, such as \( 0.4 \leq Y \leq 1 \), adjust the limits of integration accordingly: \( \int_{0.4}^1 ky(1-y) \, dy \).
This process is crucial in defining the distribution accurately, especially in deciding parameters like \( k \) that ensure the function is indeed a probability density function. Integration forms the backbone of probability calculations in continuous settings.
continuous distribution
A continuous distribution is a type of probability distribution in which a random variable can take any value within a given range. Unlike discrete distributions that deal with specific, countable outcomes, continuous distributions cover all infinitely possible values in an interval. This makes them particularly suitable for modeling real-world phenomena like height, temperature, or, in this case, the variable \( Y \).

For continuous distributions, the probability of the variable exactly equaling one specific value is zero – because there are infinitely many others it could assume. Instead, we find probabilities over a range of values, which is why integration plays a crucial role here. By integrating the pdf over a specified interval, we compute the likelihood that the variable falls within that range.

In this context:
  • The range \( 0 \leq y \leq 1 \) refers to all possible outcomes \( Y \) in the continuous distribution.
  • We use integration to ensure the total area under the pdf, \( ky(1-y) \), equals 1, confirming it’s a legitimate probability distribution.
Understanding continuous distributions is essential for interpreting data and making predictions based on models in probability and statistics, giving insights into the nuances of potential outcomes within specified ranges.

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Most popular questions from this chapter

The percentage of impurities per batch in a chemical product is a random variable \(Y\) with density function $$f(y)=\left\\{\begin{array}{ll}12 y^{2}(1-y), & 0 \leq y \leq 1 \\\0, & \text { elsewhere }\end{array}\right.$$, A batch with more than \(40 \%\) impurities cannot be sold. a. Integrate the density directly to determine the probability that a randomly selected batch cannot be sold because of excessive impurities. b. Applet Exercise Use the applet Beta Probabilities and Quantiles to find the answer to part (a).

If \(Y\) is a continuous random variable with density function \(f(y)\) that is symmetric about 0 (that is, \(f(y)=f(-y)\) for all \(y\) ) and \(E(Y)\) exists, show that \(E(Y)=0 .\) [Hint: \(E(Y)=\int_{-\infty}^{0} y f(y) d y+\int_{0}^{\infty} y f(y) d y .\) Make the change of variable \(w=-y\) in the first integral.]

The median of the distribution of a continuous random variable \(Y\) is the value \(\phi_{5}\) such that \(P\left(Y \leq \phi_{5}\right)=0.5 .\) What is the median of the uniform distribution on the interval \(\left(\theta_{1}, \theta_{2}\right) ?\)

A builder of houses needs to order some supplies that have a waiting time \(Y\) for delivery, with a continuous uniform distribution over the interval from 1 to 4 days. Because she can get by without them for 2 days, the cost of the delay is fixed at \(\$ 100\) for any waiting time up to 2 days. After 2 days, however, the cost of the delay is \(\$ 100\) plus \(\$ 20\) per day (prorated) for each additional day. That is, if the waiting time is 3.5 days, the cost of the delay is \(\$ 100+\$ 20(1.5)=\$ 130 .\) Find the expected value of the builder's cost due to waiting for supplies.

If \(Y\) is a continuous random variable such that \(E\left[(Y-a)^{2}\right]<\infty\) for all \(a,\) show that \(E\left[(Y-a)^{2}\right]\) is minimized when \(a=E(Y) .[\) Hint: \(\left.E\left[(Y-a)^{2}\right]=E\left(\\{[Y-E(Y)]+[E(Y)-a]\\}^{2}\right) .\right]\)

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