To solve this problem, we will use the standard normal distribution and the Z-score formula for both individual values and sample means.

Given:

• Mean weight, $\mu = 9$ ounces
• Standard deviation, $\sigma = 0.3$ ounces

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Part (a): Probability that a randomly selected carton has a weight greater than 9.09 ounces

We use the Z-score formula for a single data point:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 9.09$ (the value of interest)
• $\mu = 9$ (mean)
• $\sigma = 0.3$ (standard deviation)

$Z = \frac{9.09 - 9}{0.3} = \frac{0.09}{0.3} = 0.3$

Now, we need to find the probability that $Z > 0.3$. Using the standard normal distribution table (or a calculator), the cumulative probability for $Z = 0.3$ is approximately 0.6179.

Thus, the probability that $Z > 0.3$ is:

$P(Z > 0.3) = 1 - P(Z \leq 0.3) = 1 - 0.6179 = 0.3821$

So, the probability that a randomly selected carton weighs more than 9.09 ounces is 0.3821 or 38.21%.

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Part (b): Probability that the mean weight of a sample of 36 cartons is greater than 9.09 ounces

For a sample mean, the standard deviation of the sampling distribution (also called the standard error) is given by:

$\sigma_{\text{mean}} = \frac{\sigma}{\sqrt{n}}$

Where:

• $\sigma = 0.3$
• $n = 36$

$\sigma_{\text{mean}} = \frac{0.3}{\sqrt{36}} = \frac{0.3}{6} = 0.05$

Now, we calculate the Z-score for the sample mean:

$Z = \frac{\bar{X} - \mu}{\sigma_{\text{mean}}}$

Where $\bar{X} = 9.09$ (the sample mean of interest):

$Z = \frac{9.09 - 9}{0.05} = \frac{0.09}{0.05} = 1.8$

Next, using the standard normal distribution, the cumulative probability for $Z = 1.8$ is approximately 0.9641.

Thus, the probability that $Z > 1.8$ is:

$P(Z > 1.8) = 1 - P(Z \leq 1.8) = 1 - 0.9641 = 0.0359$

So, the probability that the mean weight of the sample is greater than 9.09 ounces is 0.0359 or 3.59%.

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Summary:

• (a) The probability that a randomly selected carton has a weight greater than 9.09 ounces is 38.21%.
• (b) The probability that the mean weight of a sample of 36 cartons is greater than 9.09 ounces is 3.59%.

Would you like a detailed explanation of any part? Here are five related questions to explore:

1. How does increasing the sample size affect the standard error of the mean?
2. What is the probability of a carton weighing less than 8.7 ounces?
3. How can you determine if a distribution is normally distributed based on sample data?
4. How would the probabilities change if the standard deviation were larger?
5. What is the Z-score interpretation for values below the mean?

Tip: The standard error decreases as the sample size increases, making the sample mean more reliable!