a ferry will safely accommodate 66 tons of passenger cars. assume that the mean weight of a passenger car is 1.7 tons with standard deviation 0.5 tons. if a random sample of 36 cars are loaded onto the ferry, what is the probability that the maximum safe weight will be exceeded? hint! think about the average safe weight g
The Correct Answer and Explanation is :
To determine the probability that the total weight of 36 passenger cars exceeds the ferry’s capacity of 66 tons, we first need to understand the distribution of the total weight based on the mean and standard deviation of individual car weights.
Given Information:
- Mean weight of a passenger car (μ) = 1.7 tons
- Standard deviation of the weight of a passenger car (σ) = 0.5 tons
- Number of cars (n) = 36
- Total safe weight limit of the ferry = 66 tons
Step 1: Calculate the Mean and Standard Deviation of the Sample Mean
The mean weight of the sample of 36 cars (the sample mean, ( \bar{X} )) can be calculated as follows:
[
\text{Mean of the sample} = \mu = 1.7 \text{ tons}
]
The standard deviation of the sample mean (also known as the standard error, SE) can be calculated using the formula:
[
\text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{36}} = \frac{0.5}{6} = 0.0833 \text{ tons}
]
Step 2: Calculate the Threshold Sample Mean
To find the average weight at which the ferry’s capacity is exceeded, we calculate the maximum average weight of the 36 cars that would keep the total weight within the limit:
[
\text{Maximum average weight} = \frac{66 \text{ tons}}{36} \approx 1.833 \text{ tons}
]
Step 3: Standardize the Threshold
Next, we standardize this value to find the z-score:
[
z = \frac{\bar{X} – \mu}{SE} = \frac{1.833 – 1.7}{0.0833} \approx \frac{0.133}{0.0833} \approx 1.596
]
Step 4: Find the Probability
Using the z-score, we can find the probability from the standard normal distribution table:
A z-score of 1.596 corresponds to a cumulative probability of approximately 0.9442. This indicates that about 94.42% of the time, the average weight of 36 cars will be less than 1.833 tons.
Conclusion
To find the probability that the maximum safe weight will be exceeded, we calculate:
[
P(X > 66) = 1 – P(Z < 1.596) \approx 1 – 0.9442 \approx 0.0558
]
Therefore, the probability that the maximum safe weight of 66 tons will be exceeded when loading 36 passenger cars onto the ferry is approximately 0.0558 or 5.58%. This indicates a relatively low risk of exceeding the ferry’s capacity, assuming normal distribution of car weights.