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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.

1 answer

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.