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To solve this problem, we will use the Central Limit Theorem, which states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large (usually \( n \geq 30 \)).

### Given:
- Population mean (\( \mu \)) = ₹11,200
- Population standard deviation (\( \sigma \)) = ₹3,200
- Sample size (\( n \)) = 50

We need to find the probability that the sample average monthly balance (\( \bar{x} \)) falls between ₹10,900 and ₹12,300, and the probability that it falls below ₹10,000.

#### Step 1: Standard Error of the Mean
The standard error (SE) is calculated as:
\[
SE = \frac{\sigma}{\sqrt{n}} = \frac{3200}{\sqrt{50}} \approx 452.55
\]

#### Step 2: Convert to Z-scores
To find the probabilities, we need to convert the given values to Z-scores using the formula:
\[
Z = \frac{\bar{x} - \mu}{SE}
\]

1. **For ₹10,900**:
   \[
   Z_1 = \frac{10900 - 11200}{452.55} \approx -0.66
   \]

2. **For ₹12,300**:
   \[
   Z_2 = \frac{12300 - 11200}{452.55} \approx 2.43
   \]

3. **For ₹10,000**:
   \[
   Z_3 = \frac{10000 - 11200}{452.55} \approx -2.65
   \]

#### Step 3: Find the Probabilities
Using the Z-scores, we can find the probabilities from standard normal distribution tables:

1. **Probability that \( \bar{x} \) is between ₹10,900 and ₹12,300**:
   - \( P(Z_1) \approx 0.2546 \) (area to the left of \( Z = -0.66 \))
   - \( P(Z_2) \approx 0.9927 \) (area to the left of \( Z = 2.43 \))
   - The probability that \( \bar{x} \) is between ₹10,900 and ₹12,300 is:
     \[
     P(10900 < \bar{x} < 12300) = P(Z_2) - P(Z_1) \approx 0.9927 - 0.2546 = 0.7381
     \]

2. **Probability that \( \bar{x} \) is below ₹10,000**:
   - \( P(Z_3) \approx 0.0040 \)

Therefore, the probability that the sample average monthly balance is between ₹10,900 and ₹12,300 is approximately 0.7381 (73.81%), and the probability that it is below ₹10,000 is approximately 0.0040 (0.4%).

Would you like more details or have any questions?

Here are 5 related questions for practice:
1. What is the probability that the sample average exceeds ₹12,500?
2. How would the probability change if the sample size was 100 instead of 50?
3. What is the 95% confidence interval for the sample mean given these parameters?
4. How does the Central Limit Theorem justify using a normal distribution for this problem?
5. If the population standard deviation were ₹4,000, how would that affect the probabilities?

**Tip:** Always check if the sample size is large enough to apply the Central Limit Theorem when dealing with sample means.

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 (x̄) is between ₹10,900 and ₹12,300? Below ₹10,000?

Learn how to calculate the probability that the sample average monthly balance is between ₹10,900 and ₹12,300 for a sample of 50 accounts, with a mean of ₹11,200 and standard deviation of ₹3,200, using the Central Limit Theorem and Z-scores.

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