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Magazine Chapter 17: Problem 2
The continuous random variable \(X\) has a pdf \(f(x)\) where
\(f(x)=\left\\{\begin{array}{ll}k(5-2 x) & 0 \leqslant x \leqslant 2 \\ 0 &
\text { otherwise }\end{array}\right.\)
a) Find the value of \(k\)
b) Find \(P(x>1.5)\)
c) Find \(P(0.5
Short Answer
Expert verified
a) \(k=\frac{1}{6}\); b) 0.125; c) 0.5; d) Mean: \(\frac{7}{9}\), Median: \(\approx 0.697\), Std Dev: \(\frac{2}{9}\).
Step by step solution
01
Understanding the PDF
First, we identify the form of the probability density function (pdf). Here, the function is defined as \(f(x) = k(5 - 2x)\) in the interval \(0 \leq x \leq 2\), and it is zero elsewhere. A valid pdf must integrate to 1 over its entire range.
02
Finding the Constant \(k\)
To find \(k\), we set up the equation by integrating the pdf over its range and equating it to 1:\[\int_0^2 k(5 - 2x)\, dx = 1\]Calculate:\[= k\left[5x - x^2\right]_0^2 = 1\]\(= k(10 - 4) = 1\)\(k \cdot 6 = 1\)\(k = \frac{1}{6}\).
03
Calculating \(P(x > 1.5)\)
Use the pdf to find the probability that \(x\) is greater than 1.5:\[P(x > 1.5) = \int_{1.5}^2 \frac{1}{6}(5 - 2x) \, dx\]Evaluate the integral:\[= \frac{1}{6} \left[5x - x^2\right]_{1.5}^2\]\(= \frac{1}{6}((10 - 4) - (7.5 - 2.25)) = \frac{1}{6}(6 - 5.25) = \frac{1}{6} \times 0.75 = 0.125\).
04
Calculating \(P(0.5 < x < 1.5)\)
Calculate the probability for the range:\[P(0.5 < x < 1.5) = \int_{0.5}^{1.5} \frac{1}{6}(5 - 2x) \, dx\]Perform the integral calculation:\[= \frac{1}{6} \left[5x - x^2\right]_{0.5}^{1.5}\]\(= \frac{1}{6}((7.5 - 2.25) - (2.5 - 0.25)) = \frac{1}{6}(5.25 - 2.25) = \frac{1}{6} \times 3 = 0.5\).
05
Finding the Mean
The mean \(E(X)\) of a continuous random variable is calculated as:\[E(X) = \int_{0}^{2} x \cdot \frac{1}{6}(5 - 2x) \, dx\]Solve the integral:\[= \frac{1}{6} \left[ \frac{5}{2}x^2 - \frac{2}{3}x^3 \right]_0^2\]\(= \frac{1}{6} \left(10 - \frac{16}{3}\right) = \frac{1}{6} \times \frac{14}{3} = \frac{7}{9}\).
06
Finding the Median
The median is the value of \(m\) such that \(P(X \leq m) = 0.5\). Solve:\[\int_0^m \frac{1}{6}(5 - 2x) \, dx = 0.5\]Calculate:\[= \frac{1}{6} \left[5x - x^2\right]_0^m = 0.5\]\(\frac{1}{6}(5m - m^2) = 0.5\)\(5m - m^2 = 3\)Solve: \(m^2 - 5m + 3 = 0\), find roots using quadratic formula \(m = \frac{5 \pm \sqrt{13}}{2}\). Choose the feasible \(m \approx 0.697\).
07
Finding the Standard Deviation
First, find \(E(X^2)\):\[E(X^2) = \int_{0}^{2} x^2 \cdot \frac{1}{6}(5 - 2x) \, dx\]Solve the integral:\[= \frac{1}{6} \left[\frac{5}{3}x^3 - \frac{1}{2}x^4\right]_0^2\]\(= \frac{1}{6}(\frac{40}{3} - \frac{8}{2}) = \frac{1}{6} \times \frac{8}{3} = \frac{4}{9}\).Then, calculate variance:\(Var(X) = E(X^2) - (E(X))^2 = \frac{4}{9} - \left(\frac{7}{9}\right)^2 = \frac{4}{81}\).Finally, standard deviation is \(\sqrt{Var(X)} = \frac{2}{9}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuous Random Variable
A continuous random variable is a type of random variable that can take any value within a given range. Unlike discrete random variables, which have countable outcomes, continuous random variables have an infinite number of possible values within a range.
The behavior of a continuous random variable is typically described by a probability density function (pdf). This function provides the probabilities of the variable taking on a value within a specific interval. The pdf must satisfy two conditions:
The behavior of a continuous random variable is typically described by a probability density function (pdf). This function provides the probabilities of the variable taking on a value within a specific interval. The pdf must satisfy two conditions:
- The function must be non-negative for all possible values of the random variable.
- The integral of the pdf over the entire range must equal 1, representing the total certainty that the variable will take a value in that range.
Mean and Median
When analyzing a continuous random variable, two critical measures of central tendency are the mean and median.
The **mean**, often referred to as the expected value, provides a measure of the average outcome expected over numerous repetitions of the random process. Mathematically, it is the integral of the product of the variable and its pdf over its range:\[E(X) = \int_{a}^{b} x \cdot f(x) \, dx\]In our exercise, the mean of the variable \( X \) was calculated to be \( \frac{7}{9} \).
The **median**, on the other hand, is the value that divides the probability distribution into two equal halves. To find it, you solve:\[P(X \leq m) = 0.5\]This requires finding the value of \( m \) such that the integral of the pdf from the lower bound to \( m \) equals 0.5. For the pdf in the exercise, the median was approximately \( 0.697 \).
The **mean**, often referred to as the expected value, provides a measure of the average outcome expected over numerous repetitions of the random process. Mathematically, it is the integral of the product of the variable and its pdf over its range:\[E(X) = \int_{a}^{b} x \cdot f(x) \, dx\]In our exercise, the mean of the variable \( X \) was calculated to be \( \frac{7}{9} \).
The **median**, on the other hand, is the value that divides the probability distribution into two equal halves. To find it, you solve:\[P(X \leq m) = 0.5\]This requires finding the value of \( m \) such that the integral of the pdf from the lower bound to \( m \) equals 0.5. For the pdf in the exercise, the median was approximately \( 0.697 \).
- Mean provides an average location of the distribution.
- Median gives the central point of the distribution's mass.
Standard Deviation
Standard deviation is a measure of how spread out the values of a random variable are around the mean. It's an essential tool in statistics for understanding the variability or dispersion of a dataset.
To calculate the standard deviation of a continuous random variable, you first need to determine the variance, which is the expectation of the squared deviations of the variable from its mean:\[Var(X) = E(X^2) - (E(X))^2\]Where \( E(X^2) \) is found using:\[E(X^2) = \int_{a}^{b} x^2 \cdot f(x) \, dx\]From this calculation, you can then find the standard deviation by taking the square root of the variance:\[\sigma = \sqrt{Var(X)}\]In the exercise, the variance was calculated to be \( \frac{4}{81} \), giving a standard deviation of \( \frac{2}{9} \).
To calculate the standard deviation of a continuous random variable, you first need to determine the variance, which is the expectation of the squared deviations of the variable from its mean:\[Var(X) = E(X^2) - (E(X))^2\]Where \( E(X^2) \) is found using:\[E(X^2) = \int_{a}^{b} x^2 \cdot f(x) \, dx\]From this calculation, you can then find the standard deviation by taking the square root of the variance:\[\sigma = \sqrt{Var(X)}\]In the exercise, the variance was calculated to be \( \frac{4}{81} \), giving a standard deviation of \( \frac{2}{9} \).
- Variance gives the average squared deviation from the mean.
- Standard deviation provides the average distance from the mean, adding context to the amount of variation in the data.
