Shift the mean of a (not) normal distribution but keep same scale

Question

Warning: I am not a statistician, so please go easy!

I have a (not) normally distributed measure of a population, with possible values from 0 to 100%. Its mean is 60%. I need to be able to shift the distribution to give it a mean of 70%.

I obviously can't just add 10% to all scores because anyone who previously scored 100% would now score an impossible 110%.

So is there a formula I can use to shift the population data so the mean is increased, while maintaining integrity of the limits of 0% and 100%?

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Edited because obviously this non-statistician elicited eye-rolls from the more knowledgeable people in the room.

So it's apparently not a normal distribution but it has a bell-curve shape. The rest of what I'm trying to achieve holds true so if anyone is able to offer any useful suggestions, I'd still be interested to hear.

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Editing to add further information. I have to be somewhat vague due to IP, but I'll try to explain best I can.

We have two measures that are used in a larger calculation. We know that one is inherently more trustworthy than the other, by a factor of about 10%, due to the way the data is collected. That cannot be resolved. However, when being used in the calculation, we need them to be calibrated to the same scale. The thought was that we could shift the population of one in order to match the mean of the other, so calibrate them and give us more meaningful results. Does that help?

Answer 1

Suppose your original observations are $x_1, \dots, x_N$, all between $0$ and $1$. One simple transformation would be to take an observation $x_i$ and shift it closer to $1$ by some factor $\alpha$, i.e., to reduce its distance from $1$:

$$ x_i \mapsto x_i':=1-\alpha(1-x_i). $$

If your original mean is

$$ \bar{x} = \frac{1}{N}\sum_{i=1}^N x_i, $$

then your new mean is

$$ \bar{x}' = \frac{1}{N}\sum_{i=1}^N x_i' = \frac{1}{N}\sum_{i=1}^N \big(1-\alpha(1-x_i)\big) = 1-\alpha + \alpha \bar{x}. $$

Solve this for $\alpha$,

$$ \alpha=\frac{1-\bar{x}'}{1-\bar{x}}, $$

and transform your data. Here is a little illustration in R:

xx <- c(1,rep(0.1,8),rep(0.2,25),rep(0.3,60),rep(0.4,90),rep(0.5,100),
  rep(0.6,90),rep(0.7,60),rep(0.8,25),rep(0.9,8),1)
mean(xx)
# [1] 0.5021368
desired_mean <- 0.7
alpha <- (1-desired_mean)/(1-mean(xx))
xx_prime <- 1-alpha*(1-xx)
mean(xx_prime)
0.7

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This approach will keep the order relationship of your original data, but it will change the variance. If you want to also keep the variance but change the mean (both only approximately), you could proceed as here: fit a beta distribution with parameters $a$ and $b$ to your data, then determine parameters $a'$ and $b'$ of a new beta distribution with the desired mean and variance, then transform any observation $x$ to

$$ x\mapsto F_{a',b'}^{-1}\big(F_{a,b}(x)(x)\big), $$

where $F_{a,b}$ is the cumulative distribution function (CDF) of the beta distribution. This can be done easily in R or Python. However, note that this changes the underlying fitted distribution, and the mean and variance of the transformed data will likely not precisely what you desired, but a little off.