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Chapter 17: Problem 20

The probability density function \(f(x)\), of a continuous random variable \(X\) is defined by \(f(x)=\left\\{\begin{array}{ll}\frac{1}{4} x\left(4-x^{2}\right) & 0 \leqslant x \leqslant 2 \\ 0 & \text { otherwise }\end{array}\right.\) Calculate the median value of \(X\).

Short Answer

Expert verified
The median value of \( X \) is approximately \( 1.29 \).

Step by step solution

01

Understand the PDF

The probability density function (PDF) is given as \[ f(x) = \begin{cases} \frac{1}{4} x (4 - x^2) & 0 \leq x \leq 2 \ 0 & \text{otherwise} \end{cases} \] which is a valid function over the interval [0,2] and zero elsewhere.
02

Define the Median

The median of a continuous random variable \( X \) is the value \( m \) such that the CDF \( F(m) = \int_{0}^{m} f(x) \, dx = 0.5 \). This means half the data lies below \( m \) and half above.
03

Set Up the Integral

To find the median \( m \), we need to solve one half of total probability: \[ \int_{0}^{m} \frac{1}{4} x (4 - x^2) \, dx = 0.5 \].
04

Solve the Integral

First, expand and simplify the integral: \[ \int \left( x - \frac{x^3}{4} \right) \, dx \]. The indefinite integral is \[ \int x \, dx - \frac{1}{4} \int x^3 \, dx = \frac{x^2}{2} - \frac{x^4}{16} \].
05

Apply Limits and Solve Equality

Evaluate at the limits: \[ \left[ \frac{m^2}{2} - \frac{m^4}{16} \right] - \left[ 0 \right] = 0.5 \]. Simplifying, solve \( \frac{m^2}{2} - \frac{m^4}{16} = 0.5 \).
06

Find the Roots of the Equation

First, multiply the equation by 16 to clear the fraction: \( 8m^2 - m^4 = 8 \). Rearrange into standard form for roots: \( m^4 - 8m^2 + 8 = 0 \). Let \( z = m^2 \). Solve \( z^2 - 8z + 8 = 0 \).
07

Solve the Quadratic Equation

Use the quadratic formula to solve \( z^2 - 8z + 8 = 0 \), giving roots \( z = 4 \pm \sqrt{8} \). Thus, \( z = 4 \pm 2\sqrt{2} \). Since \( z = m^2 \), choose the reasonable value \( z = 4 - 2\sqrt{2} \), so \( m = \sqrt{4 - 2\sqrt{2}} \).
08

Simplify the Median

The value \( \sqrt{4 - 2\sqrt{2}} \) is rationalized, leading to \( m \approx 1.29 \) after calculation, representing the median value.

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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, often abbreviated as PDF, is a fundamental concept in statistics and probability, especially when dealing with continuous random variables. It describes how the values of a random variable are distributed over the range of possible outcomes.

A PDF is a non-negative function, defined such that the total area under the curve over its domain is equal to 1. This ensures that the total probability, considering all possible outcomes, equals to 100%.
  • In our exercise, the PDF is defined over the interval \([0, 2]\), where \(f(x) = \frac{1}{4} x (4 - x^2)\).
  • Outside this interval, the function \(f(x)\) is zero, meaning those values have no probability of occurring.
  • The shape of the PDF helps in visualizing how the likelihood is spread across different values of the random variable.
With continuous random variables, it's important to remember that while the value of the PDF \(f(x)\) at a specific point doesn't give probability, the integral of the PDF over an interval provides probabilities.
Cumulative Distribution Function
The cumulative distribution function, or CDF, provides a convenient way to understand the probability that a random variable takes on a value less than or equal to a specified value. For a continuous random variable, the CDF is the integral of its PDF.

In mathematical terms, if \(X\) is a continuous random variable with probability density function \(f(x)\), the cumulative distribution function \(F(x)\) is given by:
  • \[ F(x) = \int_{-\infty}^{x} f(t) \, dt \]
This expression implies that the CDF at a point \(x\) is the accumulated area under the PDF curve from the leftmost limit up to \(x\).

Here's how it relates to our exercise:
  • To find the median of a continuous random variable, one can use the CDF's property that the median \(m\) satisfies \(F(m) = 0.5\).
  • In our example, solving for the median required computing the CDF by integrating \(f(x)\) from 0 to \(m\) and setting it equal to 0.5.
  • The value where the CDF equals 0.5 indicates the point at which the area under the curve is evenly divided.
Integral Calculus
Integral calculus is a major part of mathematics that deals with the accumulation of quantities and the areas under curves. In the context of probability and statistics, it is especially useful for working with continuous probability distributions.

Integrals allow us to:
  • Calculate the total probability under a PDF over a given interval.
  • Find cumulative probabilities using the CDF, as they are essentially integrals of the PDF.
  • Solve problems related to determining expected values, variances, and other statistical measures.
In our exercise, integral calculus was used to determine the CDF from the PDF, and more specifically, to solve for the median value of the continuous random variable by finding the area under the curve up to a point \(m\) where the cumulative probability is 0.5. The process involved solving an integral equation:
  • First, we set up the integral of the PDF from 0 to \(m\).
  • This integral represents the CDF \(F(m)\).
  • By solving this equation, we determined the median by ensuring that the integral equaled 0.5.
The tools of integral calculus are vital in making sense of how probabilities accumulate over the range of a random variable.

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Most popular questions from this chapter

The discrete random variable \(X\) has probability function given by $$P(x)=\left\\{\begin{array}{cc} \left(\frac{1}{4}\right)^{x-1} & x=2,3,4,5,6 \\ k & x=7 \\ 0 & \text { otherwise } \end{array}\right.$$ where \(k\) is a constant. Determine the value of \(k\) and the expected value of \(X .\)

Classify each of the following as discrete or continuous random variables. a) The number of words spelled correctly by a student on a spelling test. b) The amount of water flowing through the Niagara Falls per year. c) The length of time a student is late to class. d) The number of bacteria per cc of drinking water in Geneva. e) The amount of CO produced per litre of unleaded gas. f) The amount of a flu vaccine in a syringe. g) The heart rate of a lab mouse. h) The barometric pressure at Mount Everest. i) The distance travelled by a taxi driver per day. j) Total score of football teams in national leagues. k) Height of ocean tides on the shores of Portugal. 1) Tensile breaking strength (in newtons per square metre) of a 5 cm diameter steel cable. m) Number of overdue books in a public library.

In a large school, heights of students who are 13 years old are normally distributed with a mean of \(151 \mathrm{cm}\) and a standard deviation of \(8 \mathrm{cm} .\) Find the probability that a randomly chosen child is a) shorter than \(166 \mathrm{cm}\) b) within \(6 \mathrm{cm}\) of the average.

$$\begin{array}{l} \text { Consider the } 10 \text { data items } x_{1}, x_{2}, \ldots, x_{10} \text { . Given that } \sum_{i=1}^{10} x^{2},=1341 \text { and the } \\ \text { standard deviation is } 6.9, \text { find the value of } \bar{x} . \end{array}$$.

Medical research has shown that a certain type of chemotherapy is successful \(70 \%\) of the time when used to treat skin cancer. In a study to check the validity of such a claim, researchers chose different treatment centres and chose five of their patients at random. Here is the probability distribution of the number of successful treatments for groups of five: $$\begin{array}{|l|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & 0.002 & 0.029 & 0.132 & 0.309 & 0.360 & 0.168 \\ \hline \end{array}$$ a) Find the probability that at least two patients would benefit from the treatment. b) Find the probability that the majority of the group does not benefit from the treatment. c) Find \(E(X)\) and interpret the result. d) Show that \(\sigma(x)=1.02\) e) Graph \(P(x)\). Locate \(\mu, \mu \pm \sigma\) and \(\mu \pm 2 \sigma\) on the graph. Use the empirical rule to approximate the probability that \(x\) falls in this interval. Compare this with the actual probability.
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