Using normal tables
Tables of the standard normal distribution report values of the function Φ(z),
the cumulative distribution function (cdf) for a standard normal variable Z
(that is, E[Z] = 0, and standard deviation √V ar[Z] = 1, often denoted Z ∼
N(0,1)): Φ(z) = P[Z ≤z]. Clearly, P[Z > z] = 1 −Φ(z). Usually, only
the values corresponding to z ≥0 are reported, since, due to the symmetry of
the standard normal distribution, Φ(z) = 1 −P[Z > z] = 1 −P[Z ≤−z] =
1 −Φ(−z): if z < 0 we recover the value of Φ by calculating 1 minus the value
of Φ for the opposite (positive) of z. Our book has a full table including values
corresponding to negative z’s, so you don’t need this trick. Note that, since
the normal distribution is continuous, P[Z ≤z] = P[Z < z]. Also note that if
Z ∼N(0,1), X = σZ + μ ∼N(μ,σ). Recall that X−μ
σ ∼N(0,1).
Using a normal table (the book has one in the last pages) answer these
questions:
1. For a standard normal random variable Z (E [Z] = 0,V ar[Z] = 1, often
denoted N(0,1)) what is P[Z ≥0.6]?
2. For a (non standard) normal random variable X, such that E [X] =
0.9,s.d.(X) ≡√V ar[X] = 0.2, often denoted N(0.9,0.2)) what is P [X ≤−0.1]?
The next page has a short reminder of the use of normal tables.
Note Normal tables list the cdf of a standard normal for z listed with two dec imal paces (three significant digits), with the cdf listed to four decimal places. Rounding and interpolation may be necessary. For example, if we are looking for Φ(1.524) we see from the table that Φ(1.52) ≈ 0.9357, and Φ(1.53) ≈ 0.9370.
You could round 1.524 to 1.52 and take 0.9357 as the result. It is more common to do a linear interpolation: since 1.524 is 40% on the way be tween 1.52 and 1.53, and 0.9370 − 0.9357 = 0.0013, we can approximate
Φ(1.524) with 0.9357+0.4×0.0013 = 0.9357+0.00052 = 0.93622 ≈ 0.9362.
You will notice that, at least in this example, the added "precision" is likely to be illusory, since it is not very frequent to have probabilities nailed down to such a precision (both choices round to 93.6%). If the problem was Φ(1.525), we would interpolate with the midpoint, which is 0.93635, but, again, rounded to 93.6%. If we get closer to 1.53, the rounding would lead to 93.7% (note that the function Φ is not even close to linear, but this shows only at a higher precision). Using software, you will find that
Φ(1.524) ≈ 0.93625 ≈ 0.936, and Φ(1.525) ≈ 0.93637 ≈ 0.936 as well (moving on to Φ(1.526) we find 0.93650 ≈ 0.937), so if we are happy with three decimal places none of the above makes much of a difference.