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Let's solve this problem step by step.

### Given Information:
- Population mean (\(\mu\)) = 800 gms
- Population standard deviation (\(\sigma\)) = 140 gms
- Sample size (\(n\)) = 16

For a sample, the standard error of the mean (SEM) is calculated using the formula:
\[
\text{SEM} = \frac{\sigma}{\sqrt{n}} = \frac{140}{\sqrt{16}} = \frac{140}{4} = 35 \, \text{gms}
\]

Since the weights are normally distributed, we can use the Z-distribution to answer the questions.

---

### (a) **What is the probability that the mean of the sample would exceed 900 gms?**

We are looking for:
\[
P(\bar{X} > 900)
\]
First, calculate the Z-score for 900 gms:
\[
Z = \frac{\bar{X} - \mu}{\text{SEM}} = \frac{900 - 800}{35} = \frac{100}{35} \approx 2.857
\]
Using the Z-table, the probability for \(Z = 2.857\) is approximately 0.9979. 

The probability that the sample mean exceeds 900 gms is:
\[
P(\bar{X} > 900) = 1 - 0.9979 = 0.0021
\]
So, the probability is **0.0021** (or **0.21%**).

---

### (b) **What is the probability that the mean of the sample would be between 860 and 890 gms?**

We are looking for:
\[
P(860 < \bar{X} < 890)
\]

1. **For 860 gms**:
\[
Z = \frac{860 - 800}{35} = \frac{60}{35} \approx 1.714
\]
From the Z-table, the probability for \(Z = 1.714\) is approximately 0.9564.

2. **For 890 gms**:
\[
Z = \frac{890 - 800}{35} = \frac{90}{35} \approx 2.571
\]
From the Z-table, the probability for \(Z = 2.571\) is approximately 0.9951.

Now, the probability that the sample mean is between 860 and 890 gms:
\[
P(860 < \bar{X} < 890) = 0.9951 - 0.9564 = 0.0387
\]
So, the probability is **0.0387** (or **3.87%**).

---

### (c) **Find the limits within which 95 percent of sample means would lie on either side of the population mean.**

For 95% confidence, the Z-scores corresponding to the middle 95% of a normal distribution are approximately \(\pm 1.96\).

Now calculate the margin of error:
\[
\text{Margin of error} = Z \times \text{SEM} = 1.96 \times 35 = 68.6 \, \text{gms}
\]

Thus, the limits are:
\[
\mu \pm \text{Margin of error} = 800 \pm 68.6
\]
\[
\text{Lower limit} = 800 - 68.6 = 731.4 \, \text{gms}
\]
\[
\text{Upper limit} = 800 + 68.6 = 868.6 \, \text{gms}
\]

Therefore, **95% of the sample means** would lie between **731.4 gms** and **868.6 gms**.

---

### Recap:
- (a) Probability of exceeding 900 gms: **0.0021**
- (b) Probability of being between 860 and 890 gms: **0.0387**
- (c) 95% confidence interval for sample means: **731.4 gms to 868.6 gms**

---

Would you like any additional explanations or details on these solutions?

Here are some related questions to explore:
1. How do you interpret the result for part (a) in a practical manufacturing context?
2. What happens to the probability in part (b) if we increase the sample size to 36?
3. How does changing the population standard deviation affect the results?
4. Why is the standard error important in the context of sample means?
5. Can the sample mean ever exceed the population mean? Why or why not?

**Tip**: The Z-distribution is a powerful tool for solving problems related to sample means when the population is normally distributed!

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

Solve a statistics problem involving a manufacturing process with normally distributed item weights. Calculate probabilities for the sample mean exceeding certain values and determine the 95% confidence interval. Key concepts include the Z-Score and the Central Limit Theorem.

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