Choose a sample size from an exact normal mean distribution

For independent lifetimes $X_j\sim N(1100,70^2)$, the mean has exact distribution $\bar X\sim N(1100,4900/n)$. Thus $P(\bar X>1120)=P(Z>20\sqrt n/70)\le0.05$ requires $20\sqrt n/70\ge\Phi^{-1}(0.95)\approx1.6449$. Hence $n\ge33.1\ldots$ and the least integer is 34.

The source enters normalcdf(1120,10^99,1100,√(70²/X)), with lower 1120, upper $10^{99}$, mean 1100 and standard deviation $\sqrt{70^2/X}$. Its printed table is:

At 33 the probability exceeds 0.05; at 34 it is about 0.0478581, so 34 is the first acceptable integer. The screenshot’s surrounding values are:

Source screen values:

The population is explicitly normal, so its independent sample mean is exactly normal for every positive sample size. No CLT approximation is needed.