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 1.0000
(round to four decimal places as needed.)
b. the boat was later rated to carry only 15 passengers, and the load limit was changed to 2,580 lb. find the probability that the boat is overloaded because the mean weight of the passengers is greater than 172 (so that the total weight is greater than the maximum capacity of 2,580 lb).
the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard deviation of the sample mean

The formula for the standard deviation of the sample mean (also known as the standard error) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma$ is the population standard - deviation and $n$ is the sample size. Given $\sigma = 40.1$ lb and $n = 15$.
$\sigma_{\bar{x}}=\frac{40.1}{\sqrt{15}}\approx\frac{40.1}{3.873}\approx10.354$

Step2: Calculate the z - score

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}=172$ lb, $\mu = 177.4$ lb and $\sigma_{\bar{x}}\approx10.354$ lb.
$z=\frac{172 - 177.4}{10.354}=\frac{- 5.4}{10.354}\approx - 0.522$

Step3: Find the probability

We want to find $P(\bar{X}>172)$. Since $P(\bar{X}>172)=1 - P(\bar{X}\leq172)$, and using the standard normal distribution table for $z\approx - 0.522$, $P(Z\leq - 0.522)\approx0.3015$.
So $P(\bar{X}>172)=1 - 0.3015 = 0.6985$

Answer:

$0.6985$