https://file.aiquickdraw.com/mathbot/file/1728983568622inr6a4bv.jpg

Let's solve the problem provided in the image, which involves normal distribution and sampling.

### Given:
- The population mean \( \mu = 800 \) grams.
- The population standard deviation \( \sigma = 140 \) grams.
- Sample size \( n = 16 \).

For the sampling distribution of the sample mean, the **standard error of the mean** is calculated as:
\[
\text{SE} = \frac{\sigma}{\sqrt{n}} = \frac{140}{\sqrt{16}} = \frac{140}{4} = 35 \text{ grams}.
\]

Now, let's address each part of the question:

### Part (a): Probability that the sample mean exceeds 900 grams.
We need to find the probability that the sample mean is greater than 900 grams. This is a normal distribution problem.

1. **Z-score formula**:
\[
Z = \frac{X - \mu}{\text{SE}} = \frac{900 - 800}{35} = \frac{100}{35} \approx 2.857
\]

Using the Z-table or a standard normal distribution calculator, a Z-score of 2.857 corresponds to a cumulative probability of approximately 0.9979.

2. **To find the probability** that the sample mean exceeds 900 grams, subtract this value from 1:
\[
P(X > 900) = 1 - 0.9979 = 0.0021
\]
Thus, the probability is approximately 0.0021, or 0.21%.

### Part (b): Probability that the sample mean is between 860 and 890 grams.
We need to find \( P(860 < X < 890) \).

1. For \( X = 860 \), the Z-score is:
\[
Z = \frac{860 - 800}{35} = \frac{60}{35} \approx 1.714
\]
From the Z-table, the cumulative probability corresponding to \( Z = 1.714 \) is approximately 0.9564.

2. For \( X = 890 \), the Z-score is:
\[
Z = \frac{890 - 800}{35} = \frac{90}{35} \approx 2.571
\]
From the Z-table, the cumulative probability for \( Z = 2.571 \) is approximately 0.9949.

Now, subtract the probabilities:
\[
P(860 < X < 890) = P(Z = 2.571) - P(Z = 1.714) = 0.9949 - 0.9564 = 0.0385
\]
Thus, the probability is approximately 0.0385, or 3.85%.

### Part (c): Limits within which 95% of sample means would lie.
For a 95% confidence interval, we use the Z-value corresponding to 95% of the normal distribution, which is \( Z = 1.96 \).

The limits are given by:
\[
\mu \pm Z \times \text{SE} = 800 \pm 1.96 \times 35
\]
Calculating:
\[
800 \pm 68.6
\]
Thus, the limits are approximately \( 731.4 \text{ grams} \) and \( 868.6 \text{ grams} \).

---

Do you have any questions, or would you like further details on any part? Here are five related questions that could expand your understanding:

1. How would the results change if the sample size increased to 25?
2. What is the probability that the sample mean falls below 800 grams?
3. Can you explain how the Central Limit Theorem applies to this problem?
4. How would you calculate the probability if the population distribution was not normal?
5. What would be the 99% confidence interval for the sample mean?

**Tip**: When dealing with normal distributions, always standardize values to Z-scores to use standard normal distribution tables or calculators effectively.

A continuous manufacturing process produces items whose weights are normally distributed with a mean of 800 grams and a standard deviation of 140 grams. A random sample of 16 items is selected from this process.
a) What is the probability that the mean of the sample would exceed 900 grams?
b) What is the probability that the mean of the sample would be between 860 and 890 grams?
c) Find the limits within which 95 percent of sample means would lie on either side of the population mean.

Solve a problem involving normal distribution in manufacturing with a mean of 800 grams and a standard deviation of 140 grams. This includes calculating probabilities for sample means exceeding 900 grams, falling between 860 and 890 grams, and finding the 95% confidence interval. Suitable for students studying statistics.

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