3.4. The Normal Distribution 189xf(x)21- 3 – 2 – + + 2 + 3
Figure 3.4.1:The normal densityf(x), (3.4.6).points of inflection atx=μ±σ; see Exercise 3.4.7. By the symmetry aboutμ,it
follows that the median of a normal distribution is equal to its mean.
If we want to determineP(X≤x), then the following integration is required:P(X≤x)=∫x−∞1
√
2 πσe−(t−μ)(^2) /(2σ (^2) )
dt.
From calculus we know that the integrand does not have an antiderivative; hence,
the integration must be carried out by numerical integration procedures. The R
software uses such a procedure for its functionpnorm.IfXhas aN(μ, σ^2 ) distribu-
tion, then the R callpnorm(x, μ, σ) computesP(X≤x), whileq = qnorm(p, μ, σ)
gives thepthquantile ofX; i.e.,qsolves the equationP(X≤q)=p. We illustrate
this computation in the next example.
Example 3.4.1. Suppose the height in inches of an adult male is normally dis-
tributed with meanμ= 70 inches and standard deviationσ=4inches. Asa
graph of the pdf ofX use Figure 3.4.1 replacingμby 70 andσby 4. Suppose
we want to compute the probability that a man exceeds six feet (72 inches) in
height. Locate 72 on the figure. The desired probability is the area under the curve
over the interval (72,∞) which is computed in R by1-pnorm(72,70,4) = 0.3085;
hence, 31% of males exceed six feet in height. The 95th percentile in height is
qnorm(0.95,70,4) = 76.6inches. What percentage of males have heights within
one standard deviation of the mean? Answer:pnorm(74,70,4) - pnorm(66,70,4)
=0.6827.
Before the age of modern computing tables of probabilities for normal distribu-
tions were formulated. Due to the fact (3.4.8), only tables for the standard normal
distribution are required. LetZhave the standard normal distribution. A graph of