Question

1. checked luggage weights for delta airlines are right skewed with a population mean of 41 pounds and a standard deviation of 9 pounds. if a random sample of 45 bags is selected, what is the probability the average bag weight is more than 43 pounds? 10. er wait times at st. francis hospital are positively skewed with a mean of 73 minutes and a standard deviation of 20 minutes. a researcher selects samples of 64 patients to compare weekday vs. weekend wait times. what is the probability that the average wait time for the sample is less than 70 minutes?

Explanation:

Step1: Calculate the standard error

The formula for the standard error ($\sigma_{\bar{x}}$) is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
For problem 9: $\sigma = 9$, $n = 45$, so $\sigma_{\bar{x}}=\frac{9}{\sqrt{45}}\approx1.34$.
For problem 10: $\sigma = 20$, $n = 64$, so $\sigma_{\bar{x}}=\frac{20}{\sqrt{64}} = 2.5$.

Step2: Calculate the z - score

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
For problem 9: $\mu = 41$, $\bar{x}=43$, $z=\frac{43 - 41}{1.34}\approx1.49$.
For problem 10: $\mu = 73$, $\bar{x}=70$, $z=\frac{70 - 73}{2.5}=-1.2$.

Step3: Find the probability

For problem 9: We want $P(\bar{X}>43)$. Using the standard normal distribution table, $P(Z > 1.49)=1 - P(Z\leq1.49)$. From the table, $P(Z\leq1.49)=0.9319$, so $P(Z > 1.49)=1 - 0.9319 = 0.0681$.
For problem 10: We want $P(\bar{X}<70)$. Using the standard normal distribution table, $P(Z<-1.2)=0.1151$.

Answer:

For problem 9: The probability is approximately $0.0681$.
For problem 10: The probability is approximately $0.1151$.