Questions: 150 lb to 174 lb. Complete parts a and b below. passengers have a mean weight greater than 150 lb. The probability is (Round to four decimal places as needed.) than the maximum capacity of 2,436 lb ). The probability is . (Round to four decimal places as needed.) Do the new ratings appear to be safe when the boat is loaded with 14 passengers? Choose the correct answer below. A. Because the probability of overloading is lower with the new ratings than with the old ratings, the new ratings appear to be safe. B. Because 182.4 is greater than 174, the new ratings do not appear to be safe when the boat is loaded with 14 passengers. C. Because there is a high probability of overloading, the new ratings appear to be safe when the boat is loaded with 14 passengers. D. Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with 14 passengers.

150 lb to 174 lb. Complete parts a and b below. passengers have a mean weight greater than 150 lb. The probability is (Round to four decimal places as needed.) than the maximum capacity of 2,436 lb ). The probability is . (Round to four decimal places as needed.) Do the new ratings appear to be safe when the boat is loaded with 14 passengers? Choose the correct answer below. A. Because the probability of overloading is lower with the new ratings than with the old ratings, the new ratings appear to be safe. B. Because 182.4 is greater than 174, the new ratings do not appear to be safe when the boat is loaded with 14 passengers. C. Because there is a high probability of overloading, the new ratings appear to be safe when the boat is loaded with 14 passengers. D. Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with 14 passengers.
Transcript text: 150 lb to 174 lb. Complete parts a and b below. passengers have a mean weight greater than 150 lb. The probability is $\square$ (Round to four decimal places as needed.) than the maximum capacity of $2,436 \mathrm{lb}$ ). The probability is $\square$ $\square$. (Round to four decimal places as needed.) Do the new ratings appear to be safe when the boat is loaded with 14 passengers? Choose the correct answer below. A. Because the probability of overloading is lower with the new ratings than with the old ratings, the new ratings appear to be safe. B. Because 182.4 is greater than 174, the new ratings do not appear to be safe when the boat is loaded with 14 passengers. C. Because there is a high probability of overloading, the new ratings appear to be safe when the boat is loaded with 14 passengers. D. Because there is a high probability of overloading, the new ratings do not appear to be safe when the boat is loaded with 14 passengers.
failed

Solution

failed
failed

Solution Steps

To solve this problem, we need to calculate the probability that the total weight of 14 passengers exceeds the maximum capacity of 2,436 lb, given that each passenger has a mean weight greater than 150 lb. We will assume a normal distribution for the weights and use the given mean and standard deviation to find this probability. We will then determine if the new ratings are safe based on the calculated probability.

Step 1: Calculate Mean and Standard Deviation of Total Weight

Given:

  • Mean weight of a passenger, \( \mu = 150 \, \text{lb} \)
  • Standard deviation of weight, \( \sigma = 10 \, \text{lb} \)
  • Number of passengers, \( n = 14 \)

The mean total weight for 14 passengers is calculated as: \[ \mu_{\text{total}} = n \cdot \mu = 14 \cdot 150 = 2100 \, \text{lb} \]

The standard deviation of the total weight is: \[ \sigma_{\text{total}} = \sigma \cdot \sqrt{n} = 10 \cdot \sqrt{14} \approx 37.4166 \, \text{lb} \]

Step 2: Calculate Probability of Exceeding Maximum Capacity

The maximum capacity of the boat is \( 2436 \, \text{lb} \). We need to find the probability that the total weight exceeds this capacity. This is given by: \[ P(X > 2436) = 1 - P(X \leq 2436) = 1 - \Phi\left(\frac{2436 - \mu_{\text{total}}}{\sigma_{\text{total}}}\right) \] Where \( \Phi \) is the cumulative distribution function (CDF) of the normal distribution.

Calculating the z-score: \[ z = \frac{2436 - 2100}{37.4166} \approx 9.0000 \]

Since \( z \) is very large, \( P(X \leq 2436) \) approaches 1, leading to: \[ P(X > 2436) \approx 0.0 \]

Step 3: Determine Safety of New Ratings

Since the probability of exceeding the maximum capacity is \( 0.0 \), we conclude that the new ratings appear to be safe when the boat is loaded with 14 passengers.

Final Answer

The new ratings appear to be safe. Thus, the answer is: \[ \boxed{\text{A}} \]

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Inequality solution (2x-1)(3x-1) ≤ 7 x=
failedAnswered
Imagine Candidate Alana is running a campaign ad. In the ad, she shows images of unhoused people and people who are standing in line at food banks, and the narration talks about how Alana's opponent promotes policies that would increase poverty in the America. This is a contrast ad positive ad negative ad
failedAnswered
Figure ABCD is a kite. Find the value of x.
failedAnswered
For -y-2x=-2 determine the value of y when x=0, and the value of x when y=0. Then graph the line using the identified intercepts.
failedAnswered
(a) Compute the ratio of Total Charges (Tuition and Fees and Room and Board) for Public Four-Year In-State from 2020-21 to 2021-22 and scale the second quantity to 1. The scaled ratio is 0.981: 1.000 (Type integers or decimals rounded to three decimal places as needed.) (b) Fill in the following ratio table for 2021-22, using the values given in the second row of the table to avoid round-off error whenever possible. Private Room and Board Public Four-Year Room and Board In-State Public Two-Year Room and Board In-State 13,594 11,881 9,316 1 100 (Do not include the symbol in your answer. Do not round until the final answer. Then round to two decimal places as needed.)