Question

a boat capsized and sank in a lake. based on an assumption of a mean weight of 134 lb, the boat was rated to carry 70 passengers (so the load limit was 9,380 lb). after the boat sank, the assumed mean weight for similar boats was changed from 134 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 177.4 lb and a standard - deviation of 40.1 lb. find the probability that the boat is overloaded because the 70 passengers have a mean weight greater than 134 lb.
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Identify the sampling - distribution parameters

The sample size $n = 70$, the population mean $\mu=177.4$ lb, the population standard deviation $\sigma = 40.1$ lb. The mean of the sampling - distribution of the sample mean is $\mu_{\bar{x}}=\mu = 177.4$ lb and the standard deviation of the sampling - distribution of the sample mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{40.1}{\sqrt{70}}\approx4.797$.

Step2: Calculate the z - score

We want to find $P(\bar{X}>134)$. The z - score is calculated using the formula $z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}$. Substituting $\bar{x} = 134$, $\mu_{\bar{x}}=177.4$ and $\sigma_{\bar{x}}\approx4.797$, we get $z=\frac{134 - 177.4}{4.797}=\frac{- 43.4}{4.797}\approx - 9.05$.

Step3: Find the probability

We know that $P(\bar{X}>134)=P(Z>-9.05)$. Since the total area under the standard normal curve is 1 and $P(Z < - 9.05)\approx0$, then $P(Z>-9.05)=1 - P(Z < - 9.05)\approx1.0000$.

Answer:

$1.0000$