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Magazine Chapter 4: Problem 3
The error involved in making a certain measurement is a continuous rv \(X\) with
pdf
$$
f(x)=\left\\{\begin{array}{l}
09375\left(4-x^{2}\right) \quad-2 \leq x \leq 2 \\
\text { otherwise }
\end{array}\right.
$$
a. Sketch the graph of \(f(x)\).
b. Compute \(P(X>0)\).
c. Compute \(P(-1
Short Answer
Expert verified
P(X>0) = 5/3; P(-1.5) = 1/8.
Step by step solution
01
Understand the PDF
The given probability density function (pdf) for the random variable (rv) \( X \) is defined as \( f(x) = 0.9375(4-x^2) \) for \( -2 \leq x \leq 2 \) and 0 otherwise. The pdf is a parabolic function on the interval \( -2 \leq x \leq 2 \) and is zero outside this interval.
02
Sketch the Graph of f(x)
To sketch the graph of \( f(x) \), first note that it is a downwards-facing parabola (since the coefficient of \( x^2 \) is negative) centered at \( x = 0 \). Evaluate \( f(x) \) at key points like \( x = -2, 0, 2 \) to plot: \( f(-2) = f(2) = 0 \), and \( f(0) = 3.75 \). The graph starts and ends at \( x = -2 \) and \( x = 2 \) with value zero, peaking at \( x = 0 \).
03
Compute P(X>0)
To find \( P(X > 0) \), integrate the pdf from 0 to 2: \[P(X > 0) = \int_{0}^{2} 0.9375(4-x^2) \, dx.\]Solve the integral: \[= 0.9375 \left[ 4x - \frac{x^3}{3} \right]_{0}^{2}\]Calculating gives \[= 0.9375 \left( 8 - \frac{8}{3} \right) = 0.9375 \times \frac{16}{3} = 5 /3 .\] Therefore, \( P(X > 0) = \frac{5}{3} \).
04
Compute P(-1
To find \( P(-1 < X < 1) \), integrate the pdf from -1 to 1: \[P(-1 < X < 1) = \int_{-1}^{1} 0.9375(4-x^2) \, dx.\]Solve the integral: \[= 0.9375 \left[ 4x - \frac{x^3}{3} \right]_{-1}^{1}\]Calculate: \[= 0.9375 \left( (4 - \frac{1}{3}) - (-4 + \frac{1}{3}) \right) \]\(= 0.9375 \times \left( \frac{22}{3} \right) =\frac{3}{2}. \) Thus, \( P(-1 < X < 1) = \frac{3}{2} \).
05
Compute P(X.5)
Since \( P(X < -.5) + P(X > .5) = 1 - P(-.5 < X < .5) \), first calculate \( P(-.5 < X < .5) \) by integrating from -0.5 to 0.5: \[P(-.5 < X < .5) = \int_{-0.5}^{0.5} 0.9375(4-x^2) \, dx.\]Solve the integral:\[= 0.9375 \left[ 4x - \frac{x^3}{3} \right]_{-0.5}^{0.5}\]Evaluate: \[= 0.9375 \times 1 = \frac{15}{8}. \] Thus, \( P(X < -.5 \, \text{or} \, X > .5) = 1 - \frac{15}{8} = \frac{1}{8} \).
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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 variable that can take an infinite number of possible values within a given range. Unlike discrete random variables, which have distinct separate values, continuous random variables can vary smoothly and continuously across an interval. For example, the error measurement in the exercise follows a continuous range from -2 to 2. This means that any real number within this range can be a possible value for the error, making it a continuous random variable. The defining characteristic of a continuous random variable is its probability density function (pdf). This function determines the likelihood of the variable taking on a specific value or range of values. Because the exact probability of landing on one singular value in a continuous range is zero, the pdf helps in finding the probability over an interval. It is important to remember that the area under the pdf across its range should always equal 1, signifying the total probability.
Integral Calculus in Probability
Integral calculus plays a crucial role in determining probabilities for continuous random variables. As noted in the explanation above, when dealing with a continuous random variable, it is essential to compute the probabilities over an interval rather than a single point. This is where integration comes into play. To find the probability of a continuous random variable within a certain range, you must integrate the probability density function over that specific interval.
- For example, to find the probability of the error measurement being greater than zero, integrate the pdf from 0 to 2.
- Similarly, to determine the probability from -1 to 1, integrate the pdf over that range.
Probability Interval Calculation
Probability interval calculations are critical for understanding the likelihood of a continuous random variable falling within a specified range. Instead of using direct figures, these calculations rely on integrating the pdf over a given interval, thereby calculating probabilities based on the area under the curve, as mentioned earlier. By calculating these intervals, questions about how likely a variable is to lie within a specific range become solvable.
- For instance, to find the probability that the random variable falls between -1 and 1, the pdf is integrated over this interval to calculate the corresponding area.
- Another approach involves calculating complementary probabilities, as seen when determining the probability of the variable being outside of an interval, such as less than -0.5 or greater than 0.5.
