A ferry will safely accommodate 91 tons of passenger cars. Assume that the mean weight of a passenger car is 2.1 tons with standard deviation 0.7 tons. If a random sample of 40 cars are loaded onto the ferry, what is the probability that the maximum safe weight will be exceeded? 0.0655 © 0.0571 0.0485 0.9429
The Correct Answer and Explanation is :
To determine the probability that the total weight of 40 randomly selected passenger cars exceeds the ferry’s capacity of 91 tons, we can model the total weight as the sum of the weights of these cars. Given that the mean weight of a passenger car is 2.1 tons with a standard deviation of 0.7 tons, and assuming the weights are independent and identically distributed, we can proceed as follows:
Step 1: Calculate the mean and standard deviation of the total weight.
- Mean of total weight:
The mean total weight is the sum of the individual means:
[ \mu_{\text{total}} = 40 \times 2.1 = 84 \, \text{tons} ] - Standard deviation of total weight:
The standard deviation of the total weight is the square root of the sum of the variances (since variances add for independent variables):
[ \sigma_{\text{total}} = \sqrt{40} \times 0.7 \approx 4.429 \, \text{tons} ]
Step 2: Standardize the total weight to find the Z-score.
We want to find the probability that the total weight exceeds 91 tons. First, calculate the Z-score, which represents how many standard deviations 91 tons is above the mean:
[ Z = \frac{91 – 84}{4.429} \approx 1.58 ]
Step 3: Find the probability corresponding to the Z-score.
Using standard normal distribution tables or a calculator, we find that the cumulative probability up to a Z-score of 1.58 is approximately 0.9429.
Step 4: Calculate the probability of exceeding 91 tons.
The probability of the total weight exceeding 91 tons is the complement of the cumulative probability up to 1.58:
[ P(\text{Total weight} > 91) = 1 – 0.9429 = 0.0571 ]
Conclusion:
The probability that the total weight of 40 randomly selected passenger cars exceeds the ferry’s capacity of 91 tons is approximately 0.0571, or 5.71%.
This means that there is about a 5.71% chance that the ferry will be overloaded if 40 cars are loaded, assuming the weights of the cars are independent and identically distributed with the given mean and standard deviation.
Understanding this probability is crucial for ferry operators to ensure safety and avoid potential overloading, which could lead to structural damage or other safety hazards.
By calculating the Z-score and using the standard normal distribution, we can assess the likelihood of exceeding the weight limit and make informed decisions regarding the number of cars to load onto the ferry.
In practice, operators might use this information to set weight limits per car or implement weight checks to minimize the risk of overloading.
Additionally, this analysis assumes that the weights of the cars are normally distributed and that the sample of 40 cars is random and independent.
If these assumptions do not hold, the actual probability could differ, and alternative statistical methods might be necessary to accurately assess the risk.
Overall, this approach provides a quantitative basis for evaluating the safety of loading a specific number of cars onto the ferry.