Questions: Assume the random variable X is normally distributed, with mean μ=52 and standard deviation σ=5. Find the 10th percentile. The 10th percentile is (Round to two decimal places as needed.)

Solution Steps

The random variable \( X \) is normally distributed with the following parameters:

• Mean \( \mu = 52 \)
• Standard deviation \( \sigma = 5 \)

To find the 10th percentile, we need the z-score that corresponds to \( P(X \leq x) = 0.10 \). From standard normal distribution tables, the z-score for the 10th percentile is approximately: \[ z \approx -1.2816 \]

Using the z-score, we can convert it back to the original scale using the formula: \[ x = \mu + z \cdot \sigma \] Substituting the known values: \[ x = 52 + (-1.2816) \cdot 5 \] Calculating this gives: \[ x = 52 - 6.408 = 45.592 \] Rounding to two decimal places, we find: \[ x \approx 45.59 \]

Final Answer