Question: Chapter 5: Uniform Distribution (Concepts), Uniform Distribution Score: 9/24 Answered: 5/8 Question 6 The round off errors when measuring the distance that a long jumper has jumped is uniformly distributed between 0 and 5.6 mm. Round values to 4 decimal places when possible. a. The mean of this distribution is b. The standard deviation is c. The probability that the round off error for a jumper's distance is exactly 2.8 is P(x = 2.8) = d. The probability that the round off error for the distance that a long jumper has jumped is between 1.9 and 5.5 mm is P(1.9 < x < 5.5) = e. The probability that the jump's round off error is greater than 4.02 is P(x > 4.02) = f. P(x > 3.3 | x > 0.4) = g. Find the 54th percentile. h. Find the minimum for the upper quartile. Hint: Written Hint Helpful Videos: Probability, Conditional Probability, Conditional Probability, Percentiles The graph is a uniform distribution question with a rectangular/flat density over the interval from 0 to 5.6, positioned at center.

1. **Problem Statement:** We have a uniform distribution for round off errors between $0$ and $5.6$ mm. 2. **Uniform Distribution Basics:** For a uniform distribution $U(a,b)$: - Mean $\mu = \frac{a+b}{2}$ - Variance $\sigma^2 = \frac{(b-a)^2}{12}$ - Standard deviation $\sigma = \sqrt{\sigma^2}$ - Probability density function (pdf) $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$ - Probability for an interval $P(c < X < d) = \frac{d-c}{b-a}$ - Probability at a single point $P(X=x) = 0$ for continuous distributions - Percentile $P_p = a + p(b-a)$ where $p$ is the percentile in decimal form 3. **Given:** $a=0$, $b=5.6$ --- 4. **a. Mean:** $$\mu = \frac{0 + 5.6}{2} = \frac{5.6}{2} = 2.8$$ 5. **b. Standard Deviation:** $$\sigma = \sqrt{\frac{(5.6 - 0)^2}{12}} = \sqrt{\frac{5.6^2}{12}} = \sqrt{\frac{31.36}{12}} = \sqrt{2.6133} = 1.6166$$ 6. **c. Probability at exactly $2.8$:** For continuous uniform distribution, $$P(x=2.8) = 0$$ 7. **d. Probability between $1.9$ and $5.5$:** $$P(1.9 < x < 5.5) = \frac{5.5 - 1.9}{5.6 - 0} = \frac{3.6}{5.6} = 0.6429$$ 8. **e. Probability greater than $4.02$:** $$P(x > 4.02) = \frac{5.6 - 4.02}{5.6 - 0} = \frac{1.58}{5.6} = 0.2821$$ 9. **f. Conditional Probability $P(x > 3.3 | x > 0.4)$:** $$P(x > 3.3 | x > 0.4) = \frac{P(x > 3.3)}{P(x > 0.4)} = \frac{\frac{5.6 - 3.3}{5.6}}{\frac{5.6 - 0.4}{5.6}} = \frac{2.3/5.6}{5.2/5.6} = \frac{2.3}{5.2} = 0.4423$$ 10. **g. 54th Percentile:** $$P_{54} = 0 + 0.54 \times (5.6 - 0) = 0.54 \times 5.6 = 3.0240$$ 11. **h. Minimum for the upper quartile (75th percentile):** $$P_{75} = 0 + 0.75 \times (5.6 - 0) = 0.75 \times 5.6 = 4.2000$$ --- **Final answers:** - a. Mean = $2.8$ - b. Standard deviation = $1.6166$ - c. $P(x=2.8) = 0$ - d. $P(1.9 < x < 5.5) = 0.6429$ - e. $P(x > 4.02) = 0.2821$ - f. $P(x > 3.3 | x > 0.4) = 0.4423$ - g. 54th percentile = $3.0240$ - h. Minimum for upper quartile = $4.2000$