G-L only **Title: Understanding Probability Density Function for Bicycle Spoke Snapping**---**Introduction:**This page explores a probability density function modeling the likelihood of a bicycle spoke snapping at a given point. Let \( X \) represent the physical distance, measured in inches, from its anchoring point, where a bicycle spoke will snap. The function is defined as:\[ f(x) = \frac{6}{169} \times \left(1 - \frac{x}{13}\right) \text{ for } 0 \leq x \leq 13 \text{ and } 0 \text{ otherwise.} \]**Questions and Solutions:**a) **What is the probability that a bicycle spoke snaps less than 3 inches from its anchoring point?** - Solution: 0.135b) **What is the probability that \( X > 6 \)?** - Solution: 0.558c) **What is the probability that \( 2 < X < 8 \)?** - Solution: 0.606d) **What is the expected value of \( X \) (\( E(X) \))?** - Solution: 6.500e) **What is the expected value of \( X^2 \)?** - Solution: 50.696f) **What is the variance of \( X \)?** - Solution: 8.446g) **What is the standard deviation of \( X \)?** - Solution: 2.906h) **What is the probability that \( X \) is less than its expected value?** - Solution: 0.500i) **What is the expected value of \( X^{0.6} \)?** - Solution: [Answer not provided in the image]j) **What is the 60th percentile of \( X \)?** - Solution: [Answer not provided in the image]---**Conclusion:**This material helps in understanding the probability and statistics behind how and where a bicycle spoke might snap. Such analytical methods are invaluable in predicting failures and designing safer, more reliable components. ### Educational Exercise on Expected Values and ProbabilityIn this assignment, you will explore various statistical measures and probabilities. Please fill in the required calculations based on the data provided.1. **Expected Value of X (E(X))**: The expected value is a measure of the center of a probability distribution and gives the average outcome if the experiment were repeated many times. - Provided Expected Value: **6.500**2. **Expected Value of X²**: This is the expected value of the square of X. - Answer: **50.696**3. **Variance of X**: Variance measures how much the values of a random variable differ from the expected value. - Answer: **8.446**4. **Standard Deviation of X**: The standard deviation is the square root of the variance and provides a measure of the spread of values around the expected value. - Answer: **2.906**5. **Probability that X is less than its Expected Value**: The probability that a value is less than its expected value can be an indicator of the skewness of the distribution. - Answer: **0.500**6. **Expected Value of X^0.6**: This requires finding the expected value of X raised to the power of 0.6. - Answer: (to be filled in by the student)7. **60th Percentile of X**: This percentile indicates the value below which 60% of the data falls. - Answer: (to be filled in by the student)8. **Probability that X is within 2 Standard Deviations of its Expected Value**: This is a measure of how likely it is for the value of X to fall within a typical range around its mean. - Answer: (to be filled in by the student)9. **Probability that X = 8**: This is the probability of the random variable X equaling exactly 8. - Answer: (to be filled in by the student)10. **Comments Section**: Please add any additional observations or clarifications regarding your calculations. - Box provided for comments.### Instructions- Carefully read each question and use statistical formulas as needed to solve for the unknowns.- Input your answers in the space provided next to each question.Submit your completed answers through the provided "Submit Answer" button. Make sure to verify your calculations and ensure they logically fit within

Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a time so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts. The given density function is: fx=6169x1-x13, 0≤x≤13 It can be written as: fx=1B2, 2113x131-x13, 0≤x≤13Assume that, t=x13dtdx=113 Using the transformation method, ft=fxdxdtft=1B2, 2113x131-x1313 =1B2, 2t2-11-t2-1, 0<t<1 Thus, T~Beta(2, 2)Thus, Et=αα+β =22+2 =0.50 Vt=αβα+β2α+β+1 =2×22+222+2+1 =0.05 Thus, EX13=Et=0.50EX=13×0.50 =6.50g). The standard deviation of X can be calculated as: VX13=Vt=0.05VX=132×0.05 =8.45 SDX=VX=8.45 =2.9069h). The probability that X is less than its expected value can be calculated as: PX<EX=PX<6.50 =∫06.50fxdx =∫06.506169x1-x13dx =6169x22-x33×1306.50 =61696.5022-6.5033×13 =0.50 I). The expected value of X0.60, can be calculated as: EX0.60=∫013x0.60fxdx =∫013x0.606169x1-x13dx =6169∫013x1.60-x2.6013dx =6169x2.602.60-x3.603.60×13013 =6169132.602.60-133.603.60×13 =2.9870

Math

Statistics

Let X be physical distance, measured in inches, from its anchoring point where a bicycle spoke will snap. Then X has the following probability density function: f(x)= x(1-) for0sxs 13 and 0 otherwise. 169 a)What is the probability that a bicycle spoke snaps less than 3 inches from its anchoring point? 0.135 b) What is the probability that X >6? 0.558 c) What is the probability that 2< X < 8? 0.606 d) What is the expected value of X (E(X))? 6.500 e) What is the expected value of X2 ? 50.696