Practise interval, left-tail and central probabilities

Use mean 2 and standard deviation $\sqrt 3$: normalcdf(1.8,2.9,2,√3) gives 0.2442978362, so the answer is $\boxed{0.244}$ (3 s.f.).

Use normalcdf(−10^99,2.8,2,√3) to obtain 0.6779163887, so $P(X<2.8)=\boxed{0.678}$ (3 s.f.).

Convert $|X-2|<2\sqrt 3$ into $2-2\sqrt 3<X<2+2\sqrt 3$. Use normalcdf(2−2√3,2+2√3,2,√3) to obtain 0.9544997361, giving $\boxed{0.954}$ (3 s.f.). This agrees with the approximately 95.4% rule for two standard deviations.