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This problem is about calculating probabilities for sample means using the Central Limit Theorem (CLT). Let's break it down and solve it step by step.

### Given Data:
- Population mean (\(\mu\)) = ₹11,200
- Population standard deviation (\(\sigma\)) = ₹3,200
- Sample size (\(n\)) = 50
- We are asked to find the probability that the sample mean \(\overline{X}\) lies:
  1. Between ₹10,900 and ₹12,300
  2. Below ₹10,000

### Approach: Using the Central Limit Theorem (CLT)
According to the CLT, the sampling distribution of the sample mean \(\overline{X}\) is approximately normal if the sample size is large (which it is, \(n = 50\)). The mean of this sampling distribution is the same as the population mean (\(\mu = 11,200\)), and the standard deviation of the sample mean (standard error) is given by:

\[
\sigma_{\overline{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3,200}{\sqrt{50}} \approx 452.16
\]

Thus, the sampling distribution of \(\overline{X}\) is approximately normally distributed with:
- Mean = ₹11,200
- Standard error = ₹452.16

### 1. Probability that the sample mean is between ₹10,900 and ₹12,300

We can standardize these values into \(z\)-scores using the formula:

\[
z = \frac{\overline{X} - \mu}{\sigma_{\overline{X}}}
\]

For ₹10,900:
\[
z_{10,900} = \frac{10,900 - 11,200}{452.16} \approx -0.664
\]

For ₹12,300:
\[
z_{12,300} = \frac{12,300 - 11,200}{452.16} \approx 2.43
\]

Now, we look up these \(z\)-scores in the standard normal distribution table:
- \(P(z < -0.664) \approx 0.2532\)
- \(P(z < 2.43) \approx 0.9925\)

Thus, the probability that the sample mean is between ₹10,900 and ₹12,300 is:

\[
P(10,900 < \overline{X} < 12,300) = P(z < 2.43) - P(z < -0.664) = 0.9925 - 0.2532 = 0.7393
\]

So, the probability is approximately **0.7393** or **73.93%**.

### 2. Probability that the sample mean is below ₹10,000

For ₹10,000, the \(z\)-score is:

\[
z_{10,000} = \frac{10,000 - 11,200}{452.16} \approx -2.653
\]

Using the standard normal distribution table, we find:
- \(P(z < -2.653) \approx 0.004\)

So, the probability that the sample mean is below ₹10,000 is approximately **0.004** or **0.4%**.

### Final Answers:
1. The probability that the sample mean is between ₹10,900 and ₹12,300 is **73.93%**.
2. The probability that the sample mean is below ₹10,000 is **0.4%**.

Would you like more details or have any questions on this?

#### Related Questions:
1. How does increasing the sample size affect the standard error?
2. What is the significance of the Central Limit Theorem in this context?
3. How would the result change if the standard deviation were larger?
4. Can you explain why we use \(z\)-scores for probability calculations here?
5. What would the probability be if the sample mean needed to be between ₹10,000 and ₹12,000?

#### Tip:
When dealing with sample means, always remember that larger sample sizes reduce the variability (standard error) of the sample mean.

An auditor of a large credit card company knows that, on average, the monthly balance of any given customer is ₹11,200 with a standard deviation of ₹3,200. If he selects 50 accounts at random, what is the probability that the sample average monthly balance (X̅) is between ₹10,900 and ₹12,300? Below ₹10,000?

This problem covers probability calculations for a sample mean using the Central Limit Theorem. Given a population mean of ₹11,200 and a standard deviation of ₹3,200, we calculate the probabilities that the sample mean lies between ₹10,900 and ₹12,300, and below ₹10,000, for a sample of 50 accounts.

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