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Let \(X\) denote the temperature at which a certain chemical reaction takes place. Suppose that \(X\) has pdf $$ f(x)=\left\\{\begin{array}{cc} \frac{1}{9}\left(4-x^{2}\right) & -1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Determine the cdf and sketch it. c. Is 0 the median temperature at which the reaction takes place? If not, is the median temperature smaller or larger than 0 ? d. Suppose this reaction is independently carried out once in each of ten different labs and that the pdf of reaction time in each lab is as given. Let \(Y=\) the number among the ten labs at which the temperature exceeds 1. What kind of distribution does \(Y\) have? (Give the names and values of any parameters.)

Short Answer

Expert verified
The median temperature is greater than 0; \(Y\) is binomially distributed with parameters 10 and \(p = 0.129\).

Step by step solution

01

Analyze the PDF

The probability density function (PDF) is given as \( f(x) = \frac{1}{9}(4-x^2) \) for \( -1 \leq x \leq 2 \). This function is a downward opening quadratic function (since the coefficient of \(x^2\) is negative) with roots at \(x = \pm 2\) and vertex at \(x = 0\).
02

Sketch the PDF

The PDF \( f(x) \) is quadratic between \( -1 \) and \( 2 \) and zero outside. Sketching within \( -1 \leq x \leq 2 \), it starts at a positive value of \( \frac{5}{9} \) at \( x = -1 \), peaks at \( \frac{4}{9} \) at \( x = 0 \), and goes back to \( \frac{1}{9} \) at \( x = 2 \).
03

Determine the CDF

The cumulative distribution function (CDF), \( F(x) \), is found by integrating \( f(x) \) from \(-1\) to \(x\): \[ F(x) = \int_{-1}^{x} \frac{1}{9}(4-t^2)\,dt = \frac{1}{9}(4x - \frac{x^3}{3} + \frac{1}{3}) \]. For \( x < -1 \), \( F(x) = 0 \) and for \( x > 2 \), \( F(x) = 1 \).
04

Sketch the CDF

The CDF \( F(x) \) increases smoothly from 0 to 1 as \( x \) goes from \(-1\) to 2. It starts at 0 for \( x \leq -1 \) and climbs to 1 at \( x = 2 \). It's a cubic curve between \( -1 \) and \( 2 \).
05

Determine the Median

The median temperature is the value of \( x \) where \( F(x) = 0.5 \). Solve \( \frac{1}{9}(2x - \frac{x^3}{3} + \frac{10}{3}) = 0.5 \) for \( x \). Solving this, the median \( x \approx 0.77 \), which is greater than 0. Hence, 0 is not the median.
06

Distribution of Y

\( Y \) is the number of times the temperature exceeds 1.\( P(X > 1) = 1 - F(1) \). Integrate \( f(x) \) from 1 to 2 to find \( P(X > 1) \). \( Y \sim \text{Binomial}(10, p) \) where \( p = P(X > 1) \). Compute \( p = \frac{1}{9} \cdot \frac{7}{6} = \frac{7}{54}\).

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Key Concepts

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

Probability Density Function
The Probability Density Function (PDF) is a crucial concept in probability and statistics. It provides a way to describe the likelihood of different outcomes in a continuous random variable. In the given exercise, the PDF is represented by the function \( f(x) = \frac{1}{9}(4-x^2) \) for the interval \(-1 \leq x \leq 2\).

This function tells us how dense the probabilities are at various points within the specified interval. It's important to understand that the PDF itself does not give probabilities directly, but instead, it describes a density. To find the probability of the variable falling within a certain interval, we need to integrate the PDF over that interval.

**Key Features of a PDF:**
  • Non-negative everywhere: The function \( f(x) \geq 0 \).
  • The area under the curve equals 1, ensuring total probability sums up to 1.
  • Can represent a variety of different shapes (e.g., bell-shaped, uniform, or the quadratic form seen here).
Remember, the PDF helps us understand where outcomes are more likely and highlights regions of higher probability density.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) is closely related to the PDF but provides cumulative probabilities for a random variable. In simpler terms, it gives the probability that the random variable is less than or equal to a certain value. For the exercise in question, the CDF \( F(x) \) is calculated by integrating the PDF. This is how we move from density to actual probability.

In the exercise, the CDF is determined through the integration:
\[ F(x) = \int_{-1}^{x} \frac{1}{9}(4-t^2)\,dt = \frac{1}{9}(4x - \frac{x^3}{3} + \frac{1}{3}) \].

**Key Properties of the CDF:**
  • Monotonically non-decreasing: As \( x \) increases, \( F(x) \) never decreases.
  • For \( x < -1 \), \( F(x) = 0 \); meaning there is no probability accumulated until this point.
  • For \( x > 2 \), \( F(x) = 1 \); the point where all probability is accumulated.
  • The CDF curve provides a comprehensive view of how probability accumulates over the range, making it easier to identify median, quartiles, and other critical points.
The CDF is a helpful tool in decision-making processes where one needs to understand the probability distribution of outcomes over a range.
Binomial Distribution
The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials. In the exercise, variable \( Y \) represents such a situation where we are interested in the number of labs (out of ten) where the temperature exceeds a certain threshold.

Understanding this distribution helps in scenarios with two possible outcomes, such as success or failure, and where the events are independent of each other.

**Characteristics of the Binomial Distribution:**
  • Defined by two parameters: the number of trials \( n \) and the probability of success \( p \).
  • For each trial, there are only two outcomes (e.g., temperature exceeding 1 or not).
  • The probability of success remains constant throughout the experiments.
In this exercise, \( Y \sim \text{Binomial}(10, p) \), where \( p \) is the probability of exceeding the threshold temperature. Finding \( p \) involves integrating the PDF from 1 to 2, which leads to understanding how binomial outcomes are modeled based on the continuous PDF.
Median of a Distribution
The median of a distribution is a value that divides the probability distribution into two equal halves: 50% of the probability lies below this value, and 50% lies above. In the exercise, determining the median temperature involves finding the point where the CDF equals 0.5.

This means solving \( F(x) = 0.5 \) to find the median. From the exercise, this calculation leads to a median of \( x \approx 0.77 \), which is higher than 0, indicating that the majority of the probability mass lies to the right of zero.

**Importance of the Median:**
  • Characterizes the "center" of a distribution.
  • Less sensitive to outliers compared to the mean, making it a robust measure of central tendency.
  • Provides a simple but effective way to compare different probability distributions directly.
Knowing the median is beneficial particularly in asymmetric distributions, where it more accurately represents a central value than the mean.

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