Standard Probability Distributions - Advanced Probability Theory for Biomedical Engineers - page 24

Solution. To compute F z ( − 1 . 74), we find

F z ( − 1 . 74) = 1 − Q ( − 1 . 74) = Q (1 . 74) = 0 . 04093 ,

using (5.72) and Table A.8.

While the value a Gaussian random variable takes on is any real number between negative
infinity and positive infinity, the realistic range of values is much smaller. From Table A.9,
we note that 99.73% of the area under the curve is contained between − 3.0 and 3.0. From

the transformation z = ( x − η ) /σ , the range of values random variable x takes on is then

approximately . This notion does not imply that random variable x cannot take on a value

outside this interval, but the probability of it occurring is really very small (2 Q (3)

η ± 3 σ

= 0 . 0027).

Example 5.5.2. Suppose x is a Gaussian random variable with η = 35 and σ = 10 . Sketch the

PDF and then find P (37

≤ ≤ 51) . Indicate this probability on the sketch.

x

Solution. The PDF is essentially zero outside the interval [ 65]. The

sketch of this PDF is shown in Figure 5.10 along with the indicated probability. With

η − 3 σ, η + 3 σ ] = [5 ,

x − 35

10

z =

we have

P (37 ≤ x ≤ 51) = P (0 . 2 ≤ z ≤ 1 . 6) = F z (1 . 6) − F z (0 . 2) .

Hence P (37 ≤ x ≤ 51) = Q (0 . 2) − Q (1 . 6) = 0 . 36594 from Table A.9.

f ( α )

x

.04

.03

.02

.01

α

0 30 60

FIGURE 5.10: PDF for Example 5.5.2.