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Answer :
To determine the probability that the boat is overloaded because the 5050 passengers have a mean weight greater than 139 lb, we need to apply the concepts of normal distribution and the Central Limit Theorem.
Step-by-Step Explanation:
Identify the Given Information:
- Mean weight of a single passenger: [tex]\mu = 176.6[/tex] lb
- Standard deviation of a single passenger's weight: [tex]\sigma = 38.2[/tex] lb
- Number of passengers [tex]n = 5050[/tex]
- We want to find the probability that the mean weight of 5050 passengers is greater than 139 lb.
Calculate the Standard Error of the Mean (SEM):
The standard error of the mean is calculated using the formula:
[tex]\text{SEM} = \frac{\sigma}{\sqrt{n}}[/tex]
Substitute the given values:
[tex]\text{SEM} = \frac{38.2}{\sqrt{5050}} \approx 0.537[/tex]Determine the Z-score:
We need to find the Z-score to determine how many standard errors the sample mean of 139 lb is from the actual mean of 176.6 lb. The formula for the Z-score is:
[tex]Z = \frac{\bar{x} - \mu}{\text{SEM}}[/tex]
where [tex]\bar{x} = 139[/tex] is the sample mean.
[tex]Z = \frac{139 - 176.6}{0.537} \approx -69.86[/tex]Find the Probability:
The Z-score corresponds to a point on the standard normal distribution. Since the Z-score is negative and large in magnitude, it implies that the probability of the mean being greater than 139 lb is extremely high, essentially 1.
In practice, a Z-score so far from the mean indicates that the event of having a mean weight less than 139 lb is nearly impossible, given the provided mean and standard deviation.
In conclusion, the probability that the boat is overloaded because the 5050 passengers have a mean weight greater than 139 lb is practically 100%. This means the average weight of passengers should comfortably exceed 139 lb under normal circumstances.
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