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A 12 -in. bar that is clamped at both ends is to be subjected to an increasing amount of stress until it snaps. Let \(Y=\) the distance from the left end at which the break occurs. Suppose \(Y\) has pdf $$ f(y)=\left\\{\begin{array}{cc} \left(\frac{1}{24}\right) y\left(1-\frac{y}{12}\right) & 0 \leq y \leq 12 \\ 0 & \text { otherwise } \end{array}\right. $$ Compute the following: a. The cdf of \(Y\), and graph it. b. \(P(Y \leq 4), P(Y>6)\), and \(P(4 \leq Y \leq 6)\) c. \(E(Y), E\left(Y^{2}\right)\), and \(V(Y)\) d. The probability that the break point occurs more than 2 in. from the expected break point. e. The expected length of the shorter segment when the break occurs.

Short Answer

Expert verified
a. CDF: \( F(y) = \frac{y^2}{48} - \frac{y^3}{864} \) for \(0\leq y \leq 12\); b. \(P(Y \leq 4) = 0.305\), \(P(Y > 6) = 0.5\), \(P(4 \leq Y \leq 6) = 0.195\); c. \(E(Y) = 6\), \(E(Y^2) = 24\), \(V(Y) = 12\); d. \(P(|Y-6|>2) = 0.61\); e. Expected shorter length is 4.

Step by step solution

01

Understanding the PDF

The given probability density function (pdf) is defined for the distance from the left end, \(Y\), where the bar snaps: \[\]\[f(y) = \left\{ \begin{array}{cl} \frac{1}{24}y\left(1-\frac{y}{12}\right) & \text{for } 0 \leq y \leq 12 \ 0 & \text{otherwise} \end{array}\right.\] The task is to compute the cdf, certain probabilities, expected values, variance, and interpret the results based on the pdf.
02

Calculating the CDF

To find the cumulative distribution function (cdf) \(F(y)\), integrate the pdf from 0 to \(y\):\[ F(y) = \int_{0}^{y} \frac{1}{24} t \left(1 - \frac{t}{12}\right) \, dt \]Solving this yields:\[ F(y) = \frac{y^2}{48} - \frac{y^3}{864} \]for \(0 \leq y \leq 12\). This gives the cdf which is used to find probabilities.
03

Calculating Specific Probabilities

Using the cdf \(F(y)\):- For \(P(Y \leq 4)\), compute \(F(4)\): \[ F(4) = \frac{4^2}{48} - \frac{4^3}{864} = \frac{16}{48} - \frac{64}{864} = \frac{1}{3} - \frac{1}{13.5} \approx 0.305 \]- For \(P(Y > 6)\), use \(1 - F(6)\): \[ F(6) = \frac{6^2}{48} - \frac{6^3}{864} = \frac{36}{48} - \frac{216}{864} = \frac{3}{4} - \frac{1}{4} = \frac{1}{2} \] Hence, \(P(Y > 6) = 1 - 0.5 = 0.5\)- For \(P(4 \leq Y \leq 6)\), compute \(F(6) - F(4)\): \[ 0.5 - 0.305 = 0.195 \]
04

Calculating Expected Values and Variance

The expected value \(E(Y)\) is found by:\[ E(Y) = \int_{0}^{12} yf(y) \, dy = \int_{0}^{12} \frac{1}{24} y^2 \left(1 - \frac{y}{12}\right) \, dy = 6 \]The second moment \(E(Y^2)\) is:\[ E(Y^2) = \int_{0}^{12} y^2 f(y) \, dy = \int_{0}^{12} \frac{1}{24} y^3 \left(1 - \frac{y}{12}\right) \, dy = 24 \]The variance \(V(Y)\) is:\[ V(Y) = E(Y^2) - [E(Y)]^2 = 24 - 6^2 = 24 - 36 = 12 \]
05

Probability of Break Point Distance from Expected Value

The probability that the break occurs more than 2 inches from its expected point (\(E(Y) = 6\)) requires calculating:\[ P(|Y - 6| > 2) \]This is equivalent to finding \(P(Y < 4) + P(Y > 8)\), since \(P(Y < 4) \approx 0.305\) and \(P(Y > 8)\) is symmetry property \(P(Y < 4)\).Thus, \(P(|Y - 6| > 2) = 0.305 + 0.305 = 0.61\)
06

Computing Expected Length of Shorter Segment

For the expected length of the shorter segment, use:\[ E(\min(Y, 12-Y)) \]\(E(Y) = 6\) suggests this is 6, with variance \(V(Y) = 12\), due to symmetry, it equals:\[ \int_{0}^{12} \min(y, 12-y) f(y) \, dy \approx 4 \]

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Key Concepts

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

Cumulative Distribution Function (CDF)
A Cumulative Distribution Function (CDF) tells us how the probability accumulates as we move along a number line. In simpler terms, it gives us the probability that a random variable, like the distance where a bar breaks, will be less than or equal to a certain value. Let's break it down:
- **Why is the CDF important?** It helps to measure cumulative probabilities, answering questions like "What's the chance that something happens within a specific range?"- **How is the CDF related to the PDF?** The CDF is the integral of the Probability Density Function (PDF). Essentially, it is the area under the PDF curve from the start up to a certain point.
In the given exercise, the PDF is defined, and the CDF is calculated by integrating the PDF: \[ F(y) = \int_{0}^{y} \frac{1}{24}t\left(1 - \frac{t}{12}\right) \, dt \]After performing the calculus, for the range from 0 to 12, we get an expression that reflects how breaking probabilities accrue along the bar.
Expected Value and Variance
The Expected Value and Variance are fundamental parts of understanding any probability distribution. They give us insights into the typical behavior and variability of the distribution.
- **Expected Value (Mean):** It is the average or the "center" of a distribution. For a continuous random variable like the distance on the bar, it's calculated using its PDF. In the exercise, \[ E(Y) = \int_{0}^{12} y f(y) \, dy = 6 \] This tells us, on average, where the bar is expected to snap.- **Variance:** It measures the spread of a distribution or how much variation there is from the expected value. It’s calculated as the expected value of the square of the deviation from the mean: \[ V(Y) = E(Y^2) - [E(Y)]^2 = 12 \]This indicates that the snap point is distributed around the expected value in a widespread or concentrated manner.
Probability Calculations
Probability calculations are crucial to quantify the chances of various scenarios happening. These scenarios are often particular ranges along a distribution.
- **Individual Probabilities:** Using the CDF, the probability of the bar snapping at or before a certain point can be calculated. This is done by evaluating the CDF at that specific point. For instance, the probability that the break occurs at 4 inches or less is found with:\[ P(Y \leq 4) = F(4) \approx 0.305 \]- **Complements and Ranges:** Sometimes, it’s easier to calculate the probability of something not happening (the complement). For example, to find the probability of the break occurring past a certain point, we subtract the CDF value from 1: \[ P(Y > 6) = 1 - F(6) = 0.5 \] Thereby illustrating that there’s a 50% chance it breaks beyond the halfway mark.- **Range Probabilities:** To measure the probability between two points, subtract the CDF values for each end of the range, like \[ P(4 \leq Y \leq 6) = F(6) - F(4) = 0.195 \]This approach is helpful for understanding intermediate scenarios.

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