That sounds good to me and was leaving things until you were sorted.
I'll reply properly in the morning about the scaling as I'm still knackered, but they main thing, as you say, is to be able to sketch/get a rough idea of the distributions for different values of the parameter(s). Exponential and normal distributions are obviously related as they transform/standardise/scale the random variable then chuck it through a negative exponential. If you (OP) get your head round that, they become boringly "easy".
Edit ... I'll just try and highlight a few things, I'm not necessarily being mathematically precise, just describing a few things to reflect on.
Things like the exponential and normal pdfs for continuous random variables are obviously parameterized by things like the mean, variance, ... and an obvious thing to understanding the distributions is to understand how that works. I'm not interested in the exponential multiplier which just ensures the pdf integrates to 1, rather what the (negative) exponential exponent is. For the normal distribution, you can write it as
p(x) ~ exp(-z^2/2)
Where
z = (x-mean)/sigma.
Subtracting the mean translates the origin to be centred on the mean, and dividing by the standard deviation scales this difference, x-mean, so you can talk about being 1 stddev, 2 stddevs, ... from the mean. The standard deviation determines the effective width of the distribution as when the difference is 1 std dev from the mean
p(mean+stddev) ~ exp(-1/2) ~ 0.6
Similarly
p(mean+2*stddev) ~ exp(-4/2) ~ 0.13
p(mean+3*stddev) ~ exp(-9/2) ~ 0.01
p(mean+4*stddev) ~ exp(-16/2) ~ 0.0003
...
The pdf has decayed to ~0.01 of the centre value when you move 3 stddevs from the mean. Every further stddev you move away, the pdf decays at a faster than exponential rate, because of the squared exponent. The stddev is a scaling parameter which controls the width of the distribution.
For the exponential distribution you (OP) mentioned,
p(x) ~ exp(-x/mean)
It's an important distribution as it is memoryless and generalises the geometric distribution to continuous random variables. The mean value is the scaling parameter for the random variable x:
p(0) ~ exp(-0) = 1
p(mean) ~ exp(-1) ~ 1/3
p(2*mean) ~ exp(-2) ~ 1/9
p(3*mean) ~ exp(-3) ~ 1/27
...
Incrementing x by mean has the effect of reducing the pdf to 1/e ~ 1/3 of the previous value. It's a geometric decay. As p(5*mean) < 0.01, the effective or visible width of the pdf is [0,5*stddevs]. If couse, if you zoom in, you see the same exponential decay, as that's the geometric/memoryless property. The mean scales (determines the scale associated with) the exponential distribution.
Coming back to the original (small) error, you used
p(x) = exp(-mean*x)
Obviously this could just be a slip because of the extra parameter lambda, but such a function would basically be just wrong. Consider when the mean was large and the mean should be located near the "centre" of the distribution
p(mean) ~ exp(-mean^2) ~ 0
Which is far in the tail. The correct (one sided) exponential distribution scales the random variable so that p(mean) ~ exp(-1) and there are similar pdf areas on either side of the mean.
A bit long winded, but hopefully gets you thinking a bit about how to visualize/sketch/understand the role of the parameters for these two continuous distributions.
@tande33