Given a normally distributed random variable X with a µ=5 and σ=2, find the 30th percentile.Question 4Answera.1.24b.2.41c.3.95d.6.05

To find the 30th percentile of a normally distributed random variable, we first need to find the z-score associated with the 30th percentile. The z-score is a measure of how many standard deviations an element is from the mean.

The z-score for the 30th percentile (0.30) is approximately -0.52. This value is obtained from a standard normal distribution table or using a calculator with a normal distribution function.

Next, we use the z-score formula to find the value of X at the 30th percentile:

X = µ + Zσ

Where: µ = mean = 5 σ = standard deviation = 2 Z = z-score = -0.52

Substituting the values into the formula, we get:

X = 5 + (-0.52 * 2) = 5 - 1.04 = 3.96

So, the 30th percentile of the given normally distributed random variable X is approximately 3.96.

Therefore, the closest answer is c. 3.95.