Let Y be a continuous random variable that is normally distributed with mean, µ = 30 and standard deviation, σ = 2.4, find P( 31.2 < y < 33.6) Answer should be accurate to 4 decimal places and entered in the format 0.XXXX

To find P(31.2 < Y < 33.6), we first standardize the values using the Z-score formula:

Z = \frac{X - \mu}{\sigma}

Where X is the random variable, \mu is the mean, and \sigma is the standard deviation.

For X = 31.2:
Z_1 = \frac{31.2 - 30}{2.4} = \frac{1.2}{2.4} = 0.5

For X = 33.6:
Z_2 = \frac{33.6 - 30}{2.4} = \frac{3.6}{2.4} = 1.5

Now, we look up these Z-scores in the standard normal distribution table to find the corresponding probabilities:
P(0.5 < Z < 1.5)

Looking up in the standard normal distribution table:
P(0.5 < Z < 1.5) = P(Z < 1.5) - P(Z < 0.5)

P(Z < 1.5) = 0.9332
P(Z < 0.5) = 0.6915

P(0.5 < Z < 1.5) = 0.9332 - 0.6915 = 0.2417

Therefore,
P(31.2 < Y < 33.6) = 0.2417

\boxed{0.2417}