The pdf integrates to be 1, so I am not sure why you think it needs to be normalized? Furthermore, this gives you a continuous distribution and therefore P(x|distribution) = 0 for all x, as points have no mass. the pdf is NOT telling you a probability, but rather just the probability density function evaluation at this point. Probability is the area under the curve between two values (the limits of the integral). You could ask, "what is the probability a new observation is less than or equal to 15.2?" and the answer would be found from the cdf:
cdf(kde,15.2) % P(x <= 15.2)
ans =
0.2727
or, "What is the probability the sample will be less than 15.4 but greater than 15.2?":
cdf(kde,15.4) - cdf(kde,15.2)% P(15.2 < x < 15.4) = P(x < 15.4) - P(x < 15.2);
ans =
0.7105
I hope this helps clear this up.
