The weights of ice cream cartons are normally distributed with a mean weight of 9 ounces and a standard deviation of 0.3 ounce. ​(a) What is the probability that a randomly selected carton has a weight greater than 9.09 ​ounces? ​(b) A sample of 36 cartons is randomly selected. What is the probability that their mean weight is greater than 9.09 ​ounces?



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

---

### **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%**.

---

### **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%**.

---

### 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!

The weights of ice cream cartons are normally distributed with a mean weight of 9 ounces and a standard deviation of 0.3 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 9.09 ounces? (b) A sample of 36 cartons is randomly selected. What is the probability that their mean weight is greater than 9.09 ounces?

Learn how to calculate the probability that an ice cream carton weighs more than 9.09 ounces using normal distribution, with a mean of 9 ounces and a standard deviation of 0.3 ounce. Also, find the probability that the mean weight of a sample of 36 cartons exceeds 9.09 ounces using the central limit theorem. Suitable for students in Grades 10-12.

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