Question

Let X~N(50, 1.5). Find the number that cuts off the upper 2.5% of values of X (in other words 97.5 percentile).



Find the number that cuts off the lower 2.5% of values of X (in other words 2.5 percentile).



Notice that the area above between those two points is equal to 95% of the entire area.

Find the 95% confidence interval for the population mean
if sample mean x̄=50 and standard deviation is 1.5.


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Compare your results.

Answer 1

### Finding the Percentiles

Given that \( X \sim N(50, 1.5) \) (where 50 is the mean and 1.5 is the standard deviation):

1. **Upper 2.5% Percentile (97.5th Percentile)**:
- To find the 97.5th percentile, we can use the Z-table or standard normal distribution.
- The z-score that corresponds to the 97.5th percentile is approximately \( z = 1.96 \).
- We can use the formula for converting a z-score to a value in a normal distribution:
\[
X = \mu + z \cdot \sigma
\]
- Substituting the values:
\[
X = 50 + 1.96 \times 1.5 \approx 50 + 2.94 = 52.94
\]
- Thus, the upper 2.5% percentile (97.5th percentile) is approximately **52.94**.

2. **Lower 2.5% Percentile (2.5th Percentile)**:
- The z-score that corresponds to the 2.5th percentile is approximately \( z = -1.96 \).
- Using the same formula:
\[
X = 50 + (-1.96) \cdot 1.5 \approx 50 - 2.94 = 47.06
\]
- Thus, the lower 2.5% percentile (2.5th percentile) is approximately **47.06**.

### 95% Confidence Interval for the Population Mean

To calculate the 95% confidence interval for the population mean using a sample mean \( \bar{x} = 50 \) and standard deviation \( s = 1.5 \):

- The formula for the confidence interval is:
\[
\bar{x} \pm z \cdot \left( \frac{s}{\sqrt{n}} \right)
\]
where \( n \) is the sample size. Since we don't have a sample size, we will assume a sample size of \( n = 1 \) for this calculation.

- For a 95% confidence interval, we will use \( z = 1.96 \).

- Therefore:
\[
CI = 50 \pm 1.96 \cdot \left( \frac{1.5}{\sqrt{1}} \right)
\]
\[
CI = 50 \pm 1.96 \cdot 1.5
\]
\[
CI = 50 \pm 2.94
\]
\[
CI = [50 - 2.94, 50 + 2.94] = [47.06, 52.94]
\]

### Comparison of Results

- The numbers that cut off the upper and lower 2.5% of values (the 97.5th and 2.5th percentiles) are approximately **52.94** and **47.06**, respectively.
- The 95% confidence interval for the population mean is **[47.06, 52.94]**.

This shows that the endpoints of the 95% confidence interval for the population mean exactly match the 2.5th and 97.5th percentiles found earlier, which is consistent with the properties of the normal distribution.