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solve as in chegg first write given data then definatuion used then step bys tep uin steuial but normention step number L: is study the analyst focuses exclusively on a standardized laboratory miles-per-gallon rating measured under controlled conditions. Historical data collected from regulatory filings indicate that for this category of vehicles, the laboratory mpg ratings follow a normal distribution, though some engineers argue the shape may slightly deviate in extreme tails. After removing several outliers related to experimental engines and correcting for a clerical unit conversion error that does not affect the final analysis, the analyst determines that the distribution can be reasonably modeled with a mean of 41.8 miles per gallon and a standard deviation of 4.6 miles per gallon. Additional background information includes that previous studies examined samples of size 2, 5, 12, and 30, that the Environmental Review Board prefers probabilities expressed to four decimal places, and that an unrelated dataset concerning electric vehicle range had a mean of 312 miles, which is not relevant here but is included in the project documentation. Part A of the analysis concerns an individual vehicle selected at random from the population of all such commuter vehicles currently certified. The analyst is interested in understanding how likely it is that a randomly selected vehicle exceeds a benchmark mpg value that was proposed during a policy meeting after a lengthy discussion involving fuel taxes, consumer incentives, and manufacturing costs. Although several values were debated, including 38 mpg, 40 mpg, and 45 mpg, the final benchmark used for this probability calculation is 44 miles per gallon. Using the normal model described earlier, determine the probability that a randomly selected vehicle has a laboratory mpg rating greater than 44 mpg, taking care to clearly identify the random variable, standardize appropriately, and explain how the relevant probability is obtained from the standard normal distribution. Part B shifts focus from individual vehicles to averages, which the analyst notes is often a source of confusion among junior staff members. In this phase, three vehicles are randomly selected independently from the same population, with replacement not being relevant due to the extremely large population size. The analyst computes the average mpg of these three vehicles and compares this average to the same benchmark value discussed earlier, despite some committee members suggesting that a different cutoff should be used for averages. Ignoring those suggestions and using the same 44 mpg benchmark, determine the probability that the average mpg of the three selected vehicles exceeds 44 mpg. In your solution, you must clearly explain why the distribution used for the sample mean differs from that of a single observation, explicitly state the mean and standard deviation of the sampling distribution, and show how the probability is computed. Be careful not to confuse the standard deviation of the population with the standard deviation of the sampling distribution, as this error has appeared frequently in past submissions. Part C addresses a separate but related question that was raised during a review meeting when a stakeholder asked whether strong individual performance necessarily implies strong group performance. In response, the analyst considers the probability that all three of the randomly selected vehicles mentioned previously each individually exceed the 44 mpg benchmark. Although this question uses the same numerical cutoff as Parts A and B, it does not involve an average and should not be treated as such. Assuming independence between vehicle selections and continuing to rely on the same normal model parameters provided earlier, determine the probability that all three vehicles have mpg ratings greater than 44 mpg. Clearly explain the reasoning behind the probability structure used, and do not reuse results from Part B unless you explicitly justify why doing so would be valid, noting that in this case such reuse may lead to incorrect conclusions.

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solve as in chegg first write given data then definatuion used then step bys tep uin steuial but normention step number L: is study the analyst focuses exclusively on a standardized laboratory miles-per-gallon rating measured under controlled conditions. Historical data collected from regulatory filings indicate that for this category of vehicles, the laboratory mpg ratings follow a normal distribution, though some engineers argue the shape may slightly deviate in extreme tails. After removing several outliers related to experimental engines and correcting for a clerical unit conversion error that does not affect the final analysis, the analyst determines that the distribution can be reasonably modeled with a mean of 41.8 miles per gallon and a standard deviation of 4.6 miles per gallon. Additional background information includes that previous studies examined samples of size 2, 5, 12, and 30, that the Environmental Review Board prefers probabilities expressed to four decimal places, and that an unrelated dataset concerning electric vehicle range had a mean of 312 miles, which is not relevant here but is included in the project documentation. Part A of the analysis concerns an individual vehicle selected at random from the population of all such commuter vehicles currently certified. The analyst is interested in understanding how likely it is that a randomly selected vehicle exceeds a benchmark mpg value that was proposed during a policy meeting after a lengthy discussion involving fuel taxes, consumer incentives, and manufacturing costs. Although several values were debated, including 38 mpg, 40 mpg, and 45 mpg, the final benchmark used for this probability calculation is 44 miles per gallon. Using the normal model described earlier, determine the probability that a randomly selected vehicle has a laboratory mpg rating greater than 44 mpg, taking care to clearly identify the random variable, standardize appropriately, and explain how the relevant probability is obtained from the standard normal distribution. Part B shifts focus from individual vehicles to averages, which the analyst notes is often a source of confusion among junior staff members. In this phase, three vehicles are randomly selected independently from the same population, with replacement not being relevant due to the extremely large population size. The analyst computes the average mpg of these three vehicles and compares this average to the same benchmark value discussed earlier, despite some committee members suggesting that a different cutoff should be used for averages. Ignoring those suggestions and using the same 44 mpg benchmark, determine the probability that the average mpg of the three selected vehicles exceeds 44 mpg. In your solution, you must clearly explain why the distribution used for the sample mean differs from that of a single observation, explicitly state the mean and standard deviation of the sampling distribution, and show how the probability is computed. Be careful not to confuse the standard deviation of the population with the standard deviation of the sampling distribution, as this error has appeared frequently in past submissions. Part C addresses a separate but related question that was raised during a review meeting when a stakeholder asked whether strong individual performance necessarily implies strong group performance. In response, the analyst considers the probability that all three of the randomly selected vehicles mentioned previously each individually exceed the 44 mpg benchmark. Although this question uses the same numerical cutoff as Parts A and B, it does not involve an average and should not be treated as such. Assuming independence between vehicle selections and continuing to rely on the same normal model parameters provided earlier, determine the probability that all three vehicles have mpg ratings greater than 44 mpg. Clearly explain the reasoning behind the probability structure used, and do not reuse results from Part B unless you explicitly justify why doing so would be valid, noting that in this case such reuse may lead to incorrect conclusions.

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Asked by: nayan
Created at: 2026-02-02 21:42:21

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# Given Data - **Random variable:** Laboratory mpg rating of a commuter vehicle - **Distribution:** Normal- **Mean (\(\mu\))**: 41.8 mpg - **Standard Deviation (\(\sigma\))**: 4.6 mpg - **Benchmark:** 44 mpg - **Sample size for Part B and C:** 3 vehicles - **Probabilities required to four decimal places** - **Other data (not relevant):** Electric vehicle mean range 312 miles --- # Definitions and Concepts - **Normal Distribution:** If \( X \sim N(\mu, \sigma^2) \), then \( Z = \frac{X-\mu}{\sigma} \) follows the standard normal distribution. - **Sample Mean Distribution:** For a sample of size \( n \) from a normal population, the sample mean \( \bar{X} \) is normal with mean \( \mu \) and standard deviation \( \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} \). - **Independence:** If events are independent, the probability that all occur is the product of their individual probabilities. --- # Part A: Probability that a Random Vehicle Exceeds 44 mpg **Random Variable:** Let \( X \) = laboratory mpg rating of a randomly selected vehicle. **Distribution:** \( X \sim N(41.8, 4.6^2) \) **Goal:** Find \( P(X > 44) \) **Standardization:** \[ Z = \frac{X - \mu}{\sigma} \] \[ Z = \frac{44 - 41.8}{4.6} \approx \frac{2.2}{4.6} \approx .4783 \] **Probability Calculation:** \[ P(X > 44) = P(Z > .4783) \] Using the standard normal table or a calculator: \[ P(Z > .4783) = 1 - P(Z < .4783) \approx 1 - .6833 = .3167 \] **Final Answer (to four decimal places):** \[ \boxed{.3167} \] --- # Part B: Probability that the Average of 3 Vehicles Exceeds 44 mpg **Sample Mean (\(\bar{X}\)):** \[ \bar{X} = \text{mean of mpg ratings for 3 randomly selected vehicles} \] **Sampling Distribution:** \[ \text{Mean of } \bar{X} = \mu = 41.8 \] \[ \text{Standard deviation of } \bar{X} = \sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{4.6}{\sqrt{3}} \approx 2.6556 \] \[ \bar{X} \sim N(41.8, (2.6556)^2) \] **Standardization:** \[ Z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} \] \[ Z = \frac{44 - 41.8}{2.6556} \approx \frac{2.2}{2.6556} \approx .8286 \] **Probability Calculation:** \[ P(\bar{X} > 44) = P(Z > .8286) \] From the standard normal table: \[ P(Z < .8286) \approx .7967 \] \[ P(Z > .8286) = 1 - .7967 = .2033 \] **Final Answer (to four decimal places):** \[ \boxed{.2033} \] --- # Part C: Probability All 3 Vehicles Exceed 44 mpg (Individually) **Interpretation:** We want \( P(\text{all 3 } X_i > 44) \), where each \( X_i \) is independent and follows \( N(41.8, 4.6^2) \). From Part A, \[ P(X_i > 44) = .3167 \] **Independence:** \[ P(\text{all 3 } X_i > 44) = (.3167)^3 \approx .0318 \] **Final Answer (to four decimal places):** \[ \boxed{.0318} \] --- # Summary Table | Scenario | Probability | |----------------------------------------|-------------| | Single vehicle > 44 mpg | .3167 | | Average of 3 vehicles > 44 mpg | .2033 | | All 3 vehicles individually > 44 mpg | .0318 | --- **Note:** The calculation for the sample mean uses a reduced standard deviation (\(\sigma/\sqrt{n}\)), making the average less likely to exceed the benchmark than an individual vehicle, but more likely than all three exceeding it individually. The calculation for "all three" is not the same as for the average and should not be confused.

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