Question

a boat capsized and sank in a lake. based on an assumption of a mean weight of 133 lb, the boat was rated to carry 50 passengers (so the load limit was 6,650 lb). after the boat sank, the assumed mean weight for similar boats was changed from 133 lb to 173 lb. complete parts a and b below. a. assume that a similar boat is loaded with 50 passengers, and assume that the weights of people are normally distributed with a mean of 181.8 lb and a standard deviation of 40.3 lb. find the probability that the boat is overloaded because the 50 passengers have a mean weight greater than 133 lb. the probability is 1.0000. (round to four decimal places as needed.) b. the boat was later rated to carry only 17 passengers, and the load limit was changed to 2,941 lb. find the probability that the boat is overloaded because the mean weight of the passengers is greater than 173 (so that their total weight is greater than the maximum capacity of 2,941 lb). the probability is . (round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The standard error of the mean is given by $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 40.3$ lb and $n = 17$. So, $\sigma_{\bar{x}}=\frac{40.3}{\sqrt{17}}\approx9.787$.

Step2: Calculate the z - score

The z - score is calculated using the formula $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=173$ lb, $\mu = 181.8$ lb and $\sigma_{\bar{x}}\approx9.787$ lb. So, $z=\frac{173 - 181.8}{9.787}=\frac{- 8.8}{9.787}\approx - 0.90$.

Step3: Find the probability

We want $P(\bar{X}>173)$. Using the standard normal distribution, $P(\bar{X}>173)=1 - P(\bar{X}\leq173)$. From the z - table, $P(Z\leq - 0.90)=0.1841$. So, $P(\bar{X}>173)=1 - 0.1841 = 0.8159$.

Answer:

$0.8159$