Let p=80%=0.80, then 1-p=0.20. If n is the number of participants the variance σ2=0.80n×0.20=0.16, and standard deviation σ=√(0.16n)=0.4√n. If δ=σ=2.80, then 2.80=0.4√n, √n=2.80/0.4=7, making n=72=49. This assumes a binary (binomial) distribution with mean μ=np=49×0.80=39.2. The mean and standard deviation could be represented by μ±δ, that is, 39.2±2.80 meaning that the data probably lies between 36.4 and 42.0. This is my interpretation of the question based on the equation σ2=np(1-p) and assuming 80% is a probability (80% of the time=80% of the data). I've also assumed that the standard deviation is the margin of error, represented by δ. The sample size is sufficiently large to require no adjustment to the margin of error.