Because the standard normal random variable Z (with distribution N(0,1)) occupies a position of central importance in probability and statistics, some basic facts which are available from studying the density and the CDF of N(0,1) should be directly familiar to students, in an intuitive and hands-on manner, without having to refer to tables. Some of these facts are listed below
1. The Standard Normal Density is peaked around 0 and declines as z moves away from 0. The decline is rapid as exp)-z^2/2) goes to zero very fast. This means that observations are likely to be in a neighborhood of 0, and larger values are less likely than smaller ones, and also very large values (both negative and positive) are highly unlikely.
2. The density is symmetric around 0, since only z^2 enters the density. So the density is the same for z and -z.
3. NORMSDIST(0)=0.5000 = 1/2. This is because Z is symmetric around 0, so the set of negative values has the same probability as the set of positive values.
4. In connection with point 3 above, note that P(Z=0)=0. Like all continuous random variables, the probability of any particular outcome is 0. This seems a bit strange, that all outcomes have zero probability, but one of them must happen. This is because the probabilities are actually f(z)dz -- infinitesmal probabilities -- all of them are VERY close to zero, but not exactly 0. See the Paradox of Zero Probability for more explanation.
5. P(-0.6745<Z<+0.6745) = NORMDIST(0.6745) - NORMDIST(-0.6745) = 0.75 - 0.25 = 50%. This means that the range of values from -0.6745 to +0.6745 contains Z with 50% probability. This means that the first quartile of the standard normal density is -0.6745, the second quartile is 0, and the third quartile is +0.6745. It follows that 1.35 is the INTERQUARTILE RANGE of the standard normal distribution.
6. P(-1<Z<+1)=NORMSDIST(1)-NORMSDIST(-1).= 0.841345 - 0.158655 = 0.682689 = 68.2% This means that the chances of Z being within 1 unit of 0 are about 2/3.
7. Note that symmetry implies that F(-1) = P(X<-1) = P(X>+1) = 1 - P(X<-1)= 1 - F(1). This can easily be verified, not just for 1, but for any z and -z.
8. P(-1.65 < Z < +1.65) = NORMSDIST(1.65)-NORMSDIST(-1.65) = 0.9505 - 0.0495 = 90.1%. This means that with 90% probability, values of Z will lie between plus or minus 1.65. Only 10% of the values will lie outside this range.
9. P(-2<Z<+2)= NORMSDIST(2) - NORMSDIST(-2) = -.97725 - 0.02275 = 0.9545 = 95%. This means that 95% of the outcomes of Z lies between -2 and +2, and only 5%, or 1 in 20, fall outside this range.
10. P(-3<Z<+3) = NORMSDIST(3) - NORMSDIST(-3) = 0.99865 - 0.00135 = 99.7%. This means that in 1000 trials, only about 3 outcomes will fall outside the range -3,+3.
11. P(-5<Z<+5) = 0.9999994. Chances of Z being smaller than -5 or larger than +5 are less than 1 in a million. In 10,000 or 100,000 trials, you are unlikely to see such a large value of Z. This means that the Normal density does not generate outliers. All values are within 5 units of zero with very high probability.