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Unformatted text preview: 13 The Norma] or Gaussian Distribution The normal or Gaussian density function, which has the familiar bell—shaped curve, is omnipresent
throughout the world of statistics and probability. If X is a Gaussian random 1variable then its
probability density function fﬂx}, sometimes denoted by Nip... 02}, is
l
fxtx} = Miami} = —e‘IM’W for — oo «5. x a: no. {1.1} chE 142 Continuous Distribution Functions It may be shown that expectation is ELY] = p. and die variance Var[X] = o2. Thus the mean
and variance are the actual parameters of the distribution. The normal density function for selected
values of its parameters is shown in Figure 14. 113
ll?
illi
115
{L4
113
ll! ill u . .I  I'_ .
—I5 —4 —2 ll 2 4 ti Figure 14. The normal distribution for different values of a. As is apparent from this ﬁgure. the normal distribution is symmetric about its mean value u. It
is short and ﬂat when tr is large and tall and skinny when or is small. As we shall see momentarily,
approximately 63% of the area under fﬂx] lies within one standard deviation of its mean while
95% lies within two standard deviations. The cumulative distribution function for a random variable that is normally distributed has no
closed form, i.e., we cannot write Fxtx} in the form of an equation as we have been able to do with other distributions. This means that we must have recourse to precomputed tables. Usually
these tables provide data relating to the standard normal distribution NIIEI. I], having mean i} and variance 1. The probability density function of a standard normal random variable is obtained by
substituting the values p. = i} and o2 = 1 into Equation {7.1} . Because of its importance. and to
avoid confusion, the probability density function of a standard normal random variable X is often
denoted 43510:} and its cumulative distribution function by Ibﬂx]. Mathematical tables generally provide die values of 'DIIEJC] in increments of (101 from x = {l
[where ore] = {1.5m to J: = 3 {where 111(3) = EHQET}. Some tables include values of I up to
x = 5 (@{5} = 0.999999'3]. Since a standard normal random variable X has variance and standard
deviation equal to i, die quantity 111(1) — dirt—l} is the probability that X lies within one standard
deviation of its mean. ${2}—¢={—2] is the probability that it lies within two standard deviations of its
mean. and so on. The symmetry of the normal distribution allows us to use the standard tables to ﬁnd
¢{x} for negative arguments. For example, the symmetry of the standard normal distribution around
zero implies that Prob{X 5 —l} = P‘robL'ltr 31}andhence Pmb{X 5 —l] = l — Prob{X 51].
vaiously. this holds more generally: for any nonnegative value or. we have Prob{X 5 —:r] = 1 —Prob{X 5 or]. The values provided by tables of die standard normal distribution can be used to ﬁnd the values
of all normal distributions. Nﬂr, o 2}. Indeed, if a random variable X has the distribution Ntu. o2}.
then the random variable Z = [X — ulfo' has a standard normal distribution. To see this, observe 13 The Normal or Gaussian Distribution 143 that
Problz E z] = Probltx — Julia 5 z} = 11mlI{X E crz+rtl ”1+“ 1 '1 '1
_ E—tI—ul He J): — _m «321:3 I l
= f e‘Fﬂdt
_m 271w for t = {x — “we. Thus, if X has a normal distribution with expectation p. and standard deviation
or, then. for any constant or, Pmbll' 5 a] = PmbtiX — elftr 5 to — NIH} = Pmbtz 5 to: — siftr] In words, the probability that X, a normally distributed random variable, is less dran or is equal to
the probability that the standard normal random variable is less dran {or — “no. l[libsenre also that
if a random variable X has distribution Nut. o3}, then the derived random variable Y = 0X + b has a normal distribution with mean pry = arr + b and standard deviation of = Ialor. i.e.. Y is
More + b. [Halit Example 16 Assume that test scores in CSCST‘Q are normally distributed with mean pr = 33 and standard deviation or = 8. Since we wish to use the normal distribution, we shall treat die test scores
as real numbers rather than integer values. We wish to compute the probability of scores between
T5 and '95. w¢955X5sa=4hmr5om—PmmX5in
= Prob{Z 5 (95 — says} — Prob{Z 5 [TS — 331:3}
=nmw5Lﬂ—amm5—n
=oun_opn
= ossaz — {1158? = onus. Observe that if we were to replace '95 with 91 in the above analysis, we would effectively compute the probability of scores drat lies within one standard deviation of die mean (i.e., scores in the range
83 :I: E}. This yields PIDb{T5 5 Jr' 5 91} = Prob{Z 51}_ Prob{Z 5 _1:. =o.s413 _o.rssr = (1.5326. This result is not restricted to die example given above. Approximately two—drirds of the outcomes
associated with normally distributed random variable lie within one standard deviation of the mean. Furthermore, since @{2} — tilt—2) = (1.9"??3 — {l — GHTTS} = [1.9546, approximately 95% of the
outcomes of a normally distributed random variable lie within two standard deviations of the mean. We shall now compute the moment generating function for the normal distribution. We ﬁrst consider the standard normal case. Let the random variable Z be normally distributed with mean {1
and variance 1. Then Mzo): Ere”) = f Eezfztzla'z = f —t'_'4: 5“ 1 —tz:—Eﬁz+61—Hzlﬂ f” l —r 9): m1 2
= e dz = e z “i “i dz
[on 1.? 2H —oc 1.? 2H m l ear—zz‘ﬂdz
2H F 1
=86 ,r: J—Erﬁ—Smdz = gate f lezldz = 231”. 144 Continuous Distribution Functions Now let X be a normally distributed random variable with mean u and variance 02 and let Z be
standard normal.Then X = u+oZand Mum = Eran = Emma] = Wendi]
= ewMzrioS] = eweﬂrziaﬂlz = Eaewlalgr Let us now consider a random variable X constructed as a sum of normally distributed random
variables. Recall from Equation [5.3). that if XI. X2. . . .. X” are n independent random variables
andX=X+Xg++thhen Mﬂﬂl = Elgexlgaxz _ _ £81.] 2 Mx1iﬂlM11{3}"‘MI.{3l If these n random variables are annually distributed, with expectations n, and variances or,2 respectively. then Mﬂﬂ] = Eme+arfazrs Eme+a=§azr2 Ehawﬁsln
= E(itL+nz++.u.‘16'+licri+cr§++cr,E‘rrlr2_ Settinga = ﬂ1+ﬂ2+'”+#ﬂ ando‘2 = o2+o§+ +o";;’iI itbecomesappar‘entthatX is
normally distributed with mean u and variance o2. This result concerning sums of normally distributed random variables leads us to an important
application of the normal distribution. It turns out ﬁrst the normal distributionI in certain cases may
be used as an approximation to the discrete binomial distribution. [fthe random variable X has a
binomial distribution for which both It and die mean value np are large. then the density Function of
X is close to that of the normal distribution. In other words, the binomial distribution becomes more and more normal as n becomes large. We provide the following theorem. stated without proof. Theorem '13.] [De MoivreLaplace] Let X1. X2. .... X" be n independent andI identically
distributed Bernoulli random variables eaelr fairing the value I with probability p and the value
[1 with probabilityq = l — p. Let ” X —E[X]
X = E Xi and X“ = —.
a. «sum Then 1 b 1
lim Prober: re ct: =_ e' ﬂax.
.l'I—HJICI { _ _ ] JE [a
Observe that X is a binomial random variable and denotes the number of successes in n independent
Bernoulli trials with expectation up and standard deviation June. The random variable X‘ is the
normalized sum drat takes the values k—np
i5: . k=ﬂ,l....,n,
1.3311“! with probabilities
Prob{X* = .5} = cgpqut. s = o,1,...,n. Example I? As an exarrrple of the use of the normal distribution in approximating a binomial distribution. let us compare the results obtained for binomial distribution with n = lll and p = [1.5
and with die normal distribution with mean p. = np = 5 and o = .tnpq = 62.5. We shall just TA The Gamma Distribution 145 check a single valueI Prob{X 5 3]. For the binomial distribution we have
3
Z coo, kro.5*o.51“* = ossss.
H For the normal distribution. we obtain
Prob{X 5 S} = Prob{Z 51:3 — SLIHJZj} = Prob{Z 5 l.3'9?4] = (1.97”. The approximation is not particularly accurate. but the reason is drat H is not sufﬁciently large.
It is possible to add a (realrtirl'rJamit)I correction which attempts to adjust for the fact that we are
approximating a discrete distribution widr a continuous one. However. for small values of n it is
relatively easy to compute binomial coefﬁcients. This task becomes much harder for large :1. but it is precisely these cases that die approximation by ﬁre normal distribution beconres more accurate. One ﬁnal comment on the normal distribution: the central limit theorem. which is discussed in
Section 3.5, states that the sum of n independent random variables tends to the normal distribution
in the limit as n —> 00. This is an important result in many branches of statistics and probability
for it asserts drat no matter which distribution the in random variables have. so long as they are
independent. the distribution of their sum can be approxinrated by ﬁre nomral distribution. Thus our
previous comments concerning approximating a binomial distribution with a normal distribution should not be too surprising since a binomial random variable is a sum of n independent Bernoulli
random variables. ...
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 Spring '18
 Normal Distribution, Probability theory, coo, Julia, Halit, Pmbll

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