SPECIAL RANDOM VARIABLES
5.5 NORMAL RANDOM VARIABLES
A random variable is said to be normally distributed with parametersµandσ2, and we writeX ∼N(µ,σ2), if its density is
f(x)= 1
√2π σe−(x−µ)2/2σ2, −∞<x<∞∗
The normal densityf (x) is a bell-shaped curve that is symmetric aboutµand that attains its maximum value of 1/σ√
2π ≈0. 399/σ atx=µ(see Figure 5.7).
The normal distribution was introduced by the French mathematician Abraham de Moivre in 1733 and was used by him to approximate probabilities associated with binomial random variables when the binomial parameternis large. This result was later extended by Laplace and others and is now encompassed in a probability theorem known as the central limit theorem, which gives a theoretical base to the often noted empirical observation that, in practice, many random phenomena obey, at least approximately, a normal probability distribution. Some examples of this behavior are the height of a person, the velocity in any direction of a molecule in gas, and the error made in measuring a physical quantity.
* To verify that this is indeed a density function, see Problem 29.
The moment generating function of a normal random variable with parametersµand σ2is derived as follows:
φ(t)=E[etX] where the last equality follows since
√1
2πe−(y−tσ)2/2
is the density of a normal random variable (having parameterstσ and 1) and its integral must thus equal 1.
Upon differentiating Equation 5.5.1, we obtain φ(t)=(µ+tσ2) exp
Thusµandσ2represent respectively the mean and variance of the distribution.
An important fact about normal random variables is that ifXis normal with meanµ and varianceσ2, thenY = αX +βis normal with meanαµ+β and varianceα2σ2. That this is so can easily be seen by using moment generating functions as follows.
E[et(αX+β)] =etβE[eαtX]
=etβexp{µαt +σ2(αt)2/2} from Equation 5.5.1
=exp{(β+µα)t+α2σ2t2/2}
Because the final equation is the moment generating function of the normal random variable with meanβ+µαand varianceα2σ2, the result follows.
It follows from the foregoing that ifX ∼N(µ,σ2), then Z = X −µ
σ
is a normal random variable with mean 0 and variance 1. Such a random variable Zis said to have a standard, or unit, normal distribution. Let (·) denote its distribution function. That is,
(x)= 1
√2π x
−∞e−y2/2dy, −∞<x <∞
This result thatZ =(X −µ)/σ has a standard normal distribution whenXis normal with parameters µ andσ2 is quite important, for it enables us to write all probability statements aboutXin terms of probabilities forZ. For instance, to obtainP{X <b}, we note thatXwill be less thanbif and only if (X −µ)/σ is less than (b−µ)/σ, and so
Similarly, for anya<b,
P{a<X <b} =P
0 x
−x
P {Z < −x} P {Z > x}P
FIGURE 5.8 Standard normal probabilities.
It remains for us to compute(x). This has been accomplished by an approximation and the results are presented in Table A1 of the Appendix, which tabulates(x) (to a 4-digit level of accuracy) for a wide range of nonnegative values ofx. In addition, Program 5.5a of the text disk can be used to obtain(x).
While Table A1 tabulates(x) only for nonnegative values ofx, we can also obtain (−x) from the table by making use of the symmetry (about 0) of the standard normal probability density function. That is, forx>0, ifZrepresents a standard normal random variable, then (see Figure 5.8)
(−x)=P{Z <−x}
=P{Z >x} by symmetry
=1−(x) Thus, for instance,
P{Z <−1} =(−1)=1−(1)=1−.8413=.1587
EXAMPLE 5.5a If X is a normal random variable with mean µ = 3 and variance σ2=16, find
(a) P{X <11};
(b) P{X >−1};
(c) P{2<X <7}.
SOLUTION
(a) P{X <11} =P
$X −3
4 < 11−3 4
%
=(2)
=.9772 (b) P{X >−1} =P
$X −3
4 > −1−3 4
%
=P{Z >−1}
=P{Z <1}
=.8413 (c) P{2<X <7} =P
$2−3
4 < X −3
4 < 7−3 4
%
=(1)−(−1/4)
=(1)−(1−(1/4))
=.8413+.5987−1=.4400 ■
EXAMPLE 5.5b Suppose that a binary message — either “0” or “1” — must be transmitted by wire from location A to location B. However, the data sent over the wire are subject to a channel noise disturbance and so to reduce the possibility of error, the value 2 is sent over the wire when the message is “1” and the value−2 is sent when the message is “0.” If x,x= ±2, is the value sent at location A thenR, the value received at location B, is given byR =x+N, whereNis the channel noise disturbance. When the message is received at location B, the receiver decodes it according to the following rule:
ifR ≥.5, then “1” is concluded ifR<.5, then “0” is concluded
Because the channel noise is often normally distributed, we will determine the error probabilities whenNis a standard normal random variable.
There are two types of errors that can occur: One is that the message “1” can be incorrectly concluded to be “0” and the other that “0” is incorrectly concluded to be “1.”
The first type of error will occur if the message is “1” and 2+N <.5, whereas the second will occur if the message is “0” and−2+N ≥.5.
Hence,
P{error|message is “1”} =P{N <−1.5}
=1−(1.5)=.0668 and
P{error|message is “0”} =P{N >2.5}
=1−(2.5)=.0062 ■
EXAMPLE 5.5c The powerWdissipated in a resistor is proportional to the square of the voltageV. That is,
W =rV2
whereris a constant. Ifr =3, andVcan be assumed (to a very good approximation) to be a normal random variable with mean 6 and standard deviation 1, find
(a) E[W];
Another important result is that the sum of independent normal random variables is also a normal random variable. To see this, suppose thatXi,i=1,. . .,n, are independent, withXi being normal with meanµi and varianceσi2. The moment generating function ofn
i=1Xihas the same moment generating function as a normal random variable having meanµand varianceσ2. Hence, from the one-to-one correspondence between
moment generating functions and distributions, we can conclude thatn
i=1Xi is normal with meann
i=1µiand variancen
i=1σi2.
EXAMPLE 5.5d Data from the National Oceanic and Atmospheric Administration indicate that the yearly precipitation in Los Angeles is a normal random variable with a mean of 12.08 inches and a standard deviation of 3.1 inches.
(a) Find the probability that the total precipitation during the next 2 years will exceed 25 inches.
(b) Find the probability that next year’s precipitation will exceed that of the following year by more than 3 inches.
Assume that the precipitation totals for the next 2 years are independent.
SOLUTION LetX1andX2be the precipitation totals for the next 2 years.
(a) SinceX1+X2is normal with mean 24.16 and variance 2(3.1)2=19.22, it follows that
P{X1+X2>25} =P
$X1+X2−24.16
√19.22 > 25−24.16
√19.22
%
=P{Z >.1916}
≈.4240
(b) Since −X2 is a normal random variable with mean −12.08 and variance (−1)2(3.1)2, it follows thatX1−X2is normal with mean 0 and variance 19.22.
Hence,
P{X1>X2+3} =P{X1−X2>3}
=P
$X1−X2
√19. 22 > 3
√19. 22
%
=P{Z >.6843}
≈.2469
Thus there is a 42.4 percent chance that the total precipitation in Los Angeles during the next 2 years will exceed 25 inches, and there is a 24.69 percent chance that next year’s precipitation will exceed that of the following year by more than 3 inches. ■
Forα∈(0, 1), letzαbe such that
P{Z >zα} =1−(zα)=α
That is, the probability that a standard normal random variable is greater thanzαis equal toα(see Figure 5.9.)
1 – a a
0 za
FIGURE 5.9 P{Z >zα} =α.
The value ofzαcan, for anyα, be obtained from Table A1. For instance, since 1−(1. 645)=.05
1−(1. 96)=.025 1−(2. 33)=.01 it follows that
z.05=1.645, z.025=1.96, z.01=2.33 Program 5.5b on the text disk can also be used to obtain the value ofzα.
Since
P{Z <zα} =1−α
it follows that 100(1−α) percent of the time a standard normal random variable will be less thanzα. As a result, we callzα the 100(1−α)percentileof the standard normal distribution.