Question 8
The weights of the largemouth bass in one location are skewed right with a mean of 5.1 pounds and a standard deviation of 4.05 pounds. A fisherman looks at the weights of sample of 68 randomly selected largemouth bass that she has caught in this location.
(a) Use a normal distribution approximation to find the probability that her sample will have a mean weight of 5.7 pounds or more.
\[
P(\bar{x} \geq 5.7)=
\]
(b) Use a normal distribution approximation to find the probability her sample will have a mean weight less than 4.
\[
P(\bar{x}< 4)=
\]

Enter your answers as a decimal to 4 decimal places or as a percentage including the $\%$ symbol.
Submit

Step 1 ：Given that the weights of the largemouth bass in one location are skewed right with a mean of 5.1 pounds and a standard deviation of 4.05 pounds. A fisherman looks at the weights of sample of 68 randomly selected largemouth bass that she has caught in this location.

Step 2 ：First, we calculate the standard deviation of the sample mean, which is given by \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the standard deviation of the population and \(n\) is the sample size. Substituting the given values, we get \(\sigma_{\bar{x}} = \frac{4.05}{\sqrt{68}} \approx 0.4911\).

Step 3 ：For part (a), we want to find the probability that the sample mean is greater than or equal to 5.7 pounds. To do this, we first calculate the Z-score for 5.7, which is given by \(Z = \frac{x - \mu}{\sigma_{\bar{x}}}\), where \(x\) is the value we are interested in, \(\mu\) is the mean of the population, and \(\sigma_{\bar{x}}\) is the standard deviation of the sample mean. Substituting the given values, we get \(Z = \frac{5.7 - 5.1}{0.4911} \approx 1.2217\).

Step 4 ：The probability that a standard normal random variable is greater than or equal to this Z-score is given by \(P(Z \geq 1.2217) = 1 - P(Z < 1.2217)\). Using the standard normal distribution table or a calculator, we find that \(P(Z < 1.2217) \approx 0.8891\), so \(P(Z \geq 1.2217) = 1 - 0.8891 = 0.1109\). Therefore, the probability that the sample will have a mean weight of 5.7 pounds or more is approximately 0.1109 or 11.09\%.

Step 5 ：For part (b), we want to find the probability that the sample mean is less than 4 pounds. We follow a similar process as in part (a), but this time we calculate the Z-score for 4, which is given by \(Z = \frac{4 - 5.1}{0.4911} \approx -2.2397\).

Step 6 ：The probability that a standard normal random variable is less than this Z-score is given by \(P(Z < -2.2397)\). Using the standard normal distribution table or a calculator, we find that \(P(Z < -2.2397) \approx 0.0126\). Therefore, the probability that the sample will have a mean weight less than 4 pounds is approximately 0.0126 or 1.26\%.

Step 7 ：So, the final answers are: \(P(\bar{x} \geq 5.7) \approx \boxed{0.1109}\) and \(P(\bar{x}<4) \approx \boxed{0.0126}\).