Answer :
Final answer:
The probability that a boat with 70 passengers is overloaded due to an average weight greater than 137 lb is essentially 1 (or 100%). For a boat rated for 15 passengers with a load limit of 2,595 lb, the probability of being overloaded with an average weight greater than 173 lb per passenger is 0.7157.
Explanation:
Part A: Probability of Overload with Mean Weight Greater than 137 lb
To calculate the probability that the boat is overloaded because the 70 passengers have a mean weight greater than 137 lb, we can use the central limit theorem which states that the sampling distribution of the sample means will be normally distributed when the sample size is sufficiently large. Since there are 70 passengers, we can assume the sample size is large enough.
The problem gives us a mean of 178.9 lb and a standard deviation of 40.2 lb for the weights of people. For a sample of 70, the standard error (SE) is calculated as:
SE = σ / √ n = 40.2 lb / √ 70 ≈ 4.8 lb
We are looking for the probability that the mean weight of the 70 passengers exceeds the load limit of 137 lb per person. Since the mean weight of 137 lb is less than the given mean of 178.9 lb, this calculation will give us the probability that the boat is not overloaded. We subtract this probability from 1 to find the probability that the boat is overloaded.
Using a Z-score:
Z = (X - μ) / SE = (137 - 178.9) / 4.8 ≈ -8.73
Since this Z-score is far into the left tail of the standard normal distribution, the probability associated with this Z-score is virtually 0, meaning the probability that the boat is overloaded is essentially 1 (or 100%).
Part B: Probability of Overload with Mean Weight Greater than 173 lb
After the load limit was changed to carry only 15 passengers with a limit of 2,595 lb, we need to determine the overload probability.
The new overload threshold per passenger is 2,595 lb / 15 = 173 lb. We calculate the new SE for 15 passengers:
SE = 40.2 lb / √ 15 ≈ 10.38 lb
The Z-score is then:
Z = (173 - 178.9) / 10.38 ≈ -0.57
Using standard normal distribution tables or a calculator, we can find the probability corresponding to a Z-score of -0.57. This probability tells us the likelihood that an individual's weight is less than 173 lb. To find the probability that the mean weight is greater than 173 lb, we subtract this probability from 1.
From the Z-table, the probability associated with -0.57 is approximately 0.2843. Thus, the probability that the boat is overloaded is 1 - 0.2843 = 0.7157. After rounding to four decimal places, the probability that the boat is overloaded is 0.7157.
Final answer:
To answer part a, calculate the z-score using the formula z = (x - μ) / σ and find the probability using the standard normal table or a calculator. For part b, use the same formula to calculate the z-score and find the probability.
Explanation:
To find the probability that the boat is overloaded because the 70 passengers have a mean weight greater than 137 lb, we need to calculate the z-score for the mean weight of the passengers and then find the probability using the standard normal table or a calculator. The z-score formula is z = (x - μ) / σ, where x is the mean weight of the passengers, μ is the assumed mean weight, and σ is the standard deviation of the weight.
For part b, to find the probability that the boat is overloaded because the mean weight of the passengers is greater than 173 lb, we can calculate the z-score for the mean weight of the passengers and then find the probability using the standard normal table or a calculator. The formula for calculating the z-score is the same as in part a.
