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 .
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 40.3$ is the population standard - deviation and $n = 50$ is the sample size.
$\sigma_{\bar{x}}=\frac{40.3}{\sqrt{50}}\approx\frac{40.3}{7.071}\approx5.70$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=133$ is the sample - mean value we are interested in, and $\mu = 181.8$ is the population mean.
$z=\frac{133 - 181.8}{5.70}=\frac{-48.8}{5.70}\approx - 8.56$.

Step3: Find the probability

We want to find $P(\bar{X}>133)$. Since $P(\bar{X}>133)=1 - P(\bar{X}\leq133)$, and looking up the z - score of $z=-8.56$ in the standard normal distribution table, the value of $P(Z\leq - 8.56)\approx0$. So $P(\bar{X}>133)=1 - 0 = 1.0000$.

Answer:

$1.0000$