A boat capsized and sank in a lake. Based on an assumption of a mean weight of 142 lb , the boat was rated to carry 70 passengers (so the load limit was ). After the boat sank, the a
for similar boats was changed from 142 lb to 172 lb . Complete parts a and b below.
a. Assume that a similar boat is loaded with 70 passengers, and assume that the weights of people are normally distributed with a mean of 182.1 lb and a standard deviation of 38.6 lb . Find
boat is overloaded because the 70 passengers have a mean weight greater than 142 lb .
The probability is
(Round to four decimal places as needed.)

Question

A boat capsized and sank in a lake. Based on an assumption of a mean weight of 142 lb , the boat was rated to carry 70 passengers (so the load limit was

). After the boat sank, the a
for similar boats was changed from 142 lb to 172 lb . Complete parts a and b below.
a. Assume that a similar boat is loaded with 70 passengers, and assume that the weights of people are normally distributed with a mean of 182.1 lb and a standard deviation of 38.6 lb . Find
boat is overloaded because the 70 passengers have a mean weight greater than 142 lb .
The probability is
(Round to four decimal places as needed.)

UpStudy AI · Accepted Answer

To find the probability that the boat is overloaded because the 70 passengers have a mean weight greater than 142 lb, we need to calculate the z-score for the given mean weight and then use a standard normal distribution table or calculator to find the probability.

Given:
- Mean weight of passengers: 182.1 lb
- Standard deviation of weights: 38.6 lb
- Load limit: 9,940 lb
- Number of passengers: 70

First, let's calculate the z-score for the mean weight of 182.1 lb:
$$ z = \frac{X - \mu}{\sigma} $$
where:
- $$ X $$ is the mean weight of passengers (182.1 lb)
- $$ \mu $$ is the mean weight of passengers (182.1 lb)
- $$ \sigma $$ is the standard deviation of weights (38.6 lb)

$$ z = \frac{182.1 - 182.1}{38.6} = 0 $$

Since the z-score is 0, it means that the mean weight of passengers is exactly equal to the mean weight of 142 lb. Therefore, the boat is not overloaded based on the mean weight of passengers.

However, we need to consider the standard deviation of weights to determine if the boat is overloaded. Let's calculate the z-score for the load limit of 9,940 lb:
$$ z = \frac{X - \mu}{\sigma} $$
where:
- $$ X $$ is the load limit (9,940 lb)
- $$ \mu $$ is the mean weight of passengers (182.1 lb)
- $$ \sigma $$ is the standard deviation of weights (38.6 lb)

$$ z = \frac{9940 - 182.1}{38.6} $$

Let's calculate the z-score.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{\left(9940-182.1\right)}{38.6}$$
- step1: Remove the parentheses:
  $$\frac{9940-182.1}{38.6}$$
- step2: Subtract the numbers:
  $$\frac{9757.9}{38.6}$$
- step3: Convert the expressions:
  $$\frac{\frac{97579}{10}}{\frac{193}{5}}$$
- step4: Multiply by the reciprocal:
  $$\frac{97579}{10}\times \frac{5}{193}$$
- step5: Reduce the numbers:
  $$\frac{97579}{2}\times \frac{1}{193}$$
- step6: Multiply the fractions:
  $$\frac{97579}{2\times 193}$$
- step7: Multiply:
  $$\frac{97579}{386}$$
It seems there was an issue with the calculation. Let's try to calculate the probability manually.

The probability that the boat is overloaded can be found by calculating the area under the standard normal distribution curve to the right of the z-score 252.795337.

Since the z-score is very large, the probability of the boat being overloaded is extremely low. We can approximate this probability by considering the area under the standard normal distribution curve to the right of the z-score.

The probability that the boat is overloaded is approximately 0.
The probability that the boat is overloaded is approximately 0.

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