# Densities and the pdf [closed]

I really don't get if X is a univariate normally distributed random variable with a mean of 0 and variance of 1/16. What is the pdf of X?

and what is the State the value of this pdf at 0 as well as the probability that X = 0?

and why the answers of pdf at 0 and probability that x=0 are different?

• "what is the State the value of this pdf ..." is pretty clear a direct copy of a bookwork question ("State the value" such would be found in an assignment or similar) that has been (badly) edited (the "what is the" in front of it), presumably in an attempt to disguise its nature. Instead, state up front what the source of the question is. Please add the self-study tag, see the help center under Homework, and modify your question to clearly identify what you've done to solve the problem yourself, and indicate the specific help you need at the point you struck difficulty. – Glen_b Feb 3 '17 at 7:26
• Note that the last part of your question is a effectively a duplicate of a number of previous questions. For example answers to this one explain that density and probability are different – Glen_b Feb 3 '17 at 7:28

$$f(x)={\frac {1}{\sqrt {2\pi }\sigma}}\;e^{-{\frac {(x-\mu )^{2}}{2\sigma ^{2}}}}.$$
In your case, $\mu=0$, $\sigma^ 2=\frac{1}{16}$ so you just need to replace these values.
To find the value of the pdf at 0 you need to substitute $x$ by 0 and compute.
Now it comes the more complicated part. Because the normal distribution is continuous, it can take an infinite amount of values, from $-\infty$ to $\infty$ in a continuous way. Therefore, the probability of a single point is always zero, so $P(x=0)=0$ but for any value of $x$ this holds. This happens for all continuous distributions. Then, in the case of continuous distributions, the probability of a single point $P(x=x_0)$ does not make sense, and we always talk about probability density (or mass), $f(x_0)$.