Proving that the maximum values of these different, Normal distribution, curves are different.

Question

In a question, one variable X is Normally distributed with mean=100, variance=25 and Y is Normally distributed with mean=110, variance=36. The question asks to sketch the p.d.f of each on the same axes and the mark scheme states that the maximum value for the curve of Y should be lower than that for X. Intuitively I understand this is because the total area underneath each p.d.f must be 1 and so a greater standard deviation would mean the data is more 'spread' thus giving a lower value of the p.d.f at y=110. But, and I'm sure I'm missing something trivial, how would one prove this? I was given to understand the equation of each curve is For -inf.

Also sorry for my non existent LaTeX skills. Many thanks!

Answer 1

Hint: The maximum pdf value occurs at the mean (which you can show, or perhaps just assume). So can you just inspect the pdf functions of normal distributions assuming you are at the mean, and make a comparison of pdf values in terms of standard deviation? If you don't know the pdf formula for a normal distribution there's basically no way to proceed here.

Answer 2

Your intuitive reasoning is accurate. Intuition is very important here, but let us calculate.

The density function for $X$ is $\frac{1}{6\sqrt{2\pi}}e^{-(x-100)^2/72}$. The maximum of this is at $x=100$. The maximum value is $\frac{1}{6\sqrt{2\pi}}$. Do a similar calculation for the maximum of the density function of $Y$.