Sec. 22.8 Normal Distribution 1085

(c) Using (b). prove (3).(d) Prove (4).(e) Show that the Poisson distribution has the moment generating function G(t) and prove (6),

where

G(t) = e-"e,.e'.

(f) Prove x (~) = M (~ = :). Using this, prove (9).

15. By definition, the multinomial distribution has the probability function

where 0 ~ Xj ~ n. j = 1,. . . ,k, and Xl + . . . + X/c = n; also, PI + P2 + . . . + P.. a 1.Show that this is the probability of obtaining in n independent trials precisely Xj AJ's.j = I,. . . ,k, where Pj is the probability of Aj in a single trial.

22.8 Normal DistributionTurning from discrete to continuous distributions, in this section we discuss the normaldistribution. This is the most important continuous distribution because in applicationsmany random variables are normal random variables (that is, they have a normaldistribution) or they are approximately normal or can be transformed into normal randomvariables in a relatively simple fashion. Furthermore, the normal distribution is a usefulapproximation of more complicated distributions, and it also occurs in the proofs of various

statistical tests.

The normaldensity

(1)

where exp is the exponential function with base e = 2.718 . . '. This is simpler than it

may at flfSt look. f(x) has these features (see also Fig. 487 on the next page).

1. JL is the mean and 0" the standard deviation.

2. 1/(0"\12;) is a constant factor that makes the area under the curve equal to I, as itmust be by (10), Sec. 22.5.

3. The curve of f(x) is symmetric with respect to x = JL because the exponent isquadratic. Hence for JL = 0 it is symmetric with respect to the y-axis x = 0 (Fig. 487,

"bell-shaped curves ").

4. The exponential function in (1) goes to zero very fast-the faster the smaller thestandard deviation 0" is, as it should be (Fig. 487).

n!f(Xl' . . . ,.1",,) = 1'\*1 . . . Pk:'.Xt!.. .Xk!

or Gauss distribution is defined as the distribution with the

f(x) = ~CIP [ - +(9-)2] (C1' > 0)

-1086 Data Analysis. Probability Theory ChIP. 22

.

Distribution Function F(x)From (7) in Sec. 22.5 and (1) we see that the normal distribution has the

function

(2)

Here we needed x as the upper limit of integration and wrote v (instead of x) in the

integrand.For the corresponding standardized normal disbibution with mean 0 and standard

deviation I we denote F(x) by 4>(z). Then we simply have from (2)

(3)

This integral cannot be integrated by one of the methods of calculus. But this is no serioushandicap because the integral has been tabulated (Table A 7 in Appendix 5) since oneneeds its values in working with the normal distribution. The curve of tI>(z) is S-shaped.It increases monotone (why?) from 0 to I and intersects the vertical axis at 1/2 (why?),as shown in Fig. 488.

It is now of greatest practical importance that the general F(x) in (2) with any I.L andu can be expressed in tenDs of the tabulated standard tI>(z):

~. .'~.-,.. ~..

~

with IA. = 0Fig. 487. Density (1) of the normal distributionfor various values of 0'

distribution

Distribution function cI>(z) of the normal distributionwith mean 0 and variance 1

Fig. 488.

Sec. 22.8

(Use of the normc

The di.5tribution furrelated to the stand

THEOREM 1

(4)

Comparing (2) and (3), we see that we should setPROOF.

Probabilities corresponding to intervals will be needed quite frequently in statistics in

Chap. 23.

(Normal probabilities for intervals)The probability that a normal random variable X with mean JL and standard deviation 0"

assume any value in an interval a < x ~ b is

THEOREM 2

(5)

Formula (2) in Sec. 22.5 gives the fIrSt equality in (5), and (4) in this section gives thesecond equality. -04

PROOF.

Numerical Values~ :. In practical work with the normal distribution it is good to remember that about 2/3 of.~.,... ~.. . .all values of X to be observed will lie between /.l. :t 0", about 95% between /.l. :t 20", and

practically all between the three-sigma limits /.l. :t 30". More precisely, by Table A 7 in

Appendix 5

(a) P(/.l. - 0" < X ~ /.l. + 0') - 68*

(6) (b) ~1£-1~< X S p, + 2"'- 95;8. (c) P{p, - 3l1 < X ~ II. + 30') - 99.7f11.

These formulas are illustrated in Fig. 489 on the next page.

t,jj

IiL :

'tl-

i'. i

jI

-.. :11

II~, !;, Ii

IiII

!!!

I

'1

;j

!I:, f;

!

j,

:i!

x-p.Then v = x gives " -0'

e-u2/2 udu = "'(7) . ~

These are obtained as follows.

I -

1088 Data Analysis. Probability Theory Chap. 22

The formulas in (6) show that a value deviating from IJ. by ~ than 0'. 20'. <X' 30' willoccur in one of about 3. 20. and 300 mals. respectively.

#1-0' #I #1+0'

In tests (Chap. 23) we shall ask conversely for the intervals that correspond to certaingiven probabilities; practically most important are probabilities of 95%, 99%, and 99.9%.For these, Table AS in Appendix 5 gives the answers II. + 20'. II. ::t 2.50'. and II. ::t 3.30',respectively. More precisely,

(8)

(b)

(C)

(7)

Working with the Normal Tables in Appendix 5There are two normal tables in Appendix 5, Tables A 7 and A8. If you want probabilities,use Table A 7. If probabilities are given and corresponding intervals or x-values are wanted,use Table A8. The following examples are typical. Do them with care, verifing all values,and don't just regard them as dull exercises for your software. Make sketches of the densityto see whether the results look reasonable.

EXAMPLE 1 Reading entries from Table A7If X is standardized nonnaI (so chit Ii 8 0, 0' - I), then

- . .. .

'~.-,.. ~~ .". . p(X ~ 2.44) - 0.9927 - 991"

P(X ~ -1.16) a 0.12:.) - 12'"

P(X ~ I) m I - P(X ~ I) a I - 0.8413 - 0.1587 by (7). Sec. 22.3

P(I.O ~ X ~ 1.1) - ~I.I) - ~I.O) - 0.9641 - 0.8413 - 0.1221.

EXAMPLE 2 ". Probabilities for given intervals, Table A7Let X be normal with mean 0.8 and variance 4 (so that tr 8 2). 11Ieft by (4) and (5)

(2.44 - 0.8))P(X ~ 2.44) = F(2.44) E 4> 2 = 41(0.82) = 0.7939 - ~

or if you like it better (similarly in the odIer cases)

-

Sec.

lK 2.25.-.2.25-..

p-2<7 "(6)(0)

Illustration of formula (6)Fig. 489.

..

Sec. 22.8

EXAMPLE 3 Unknown values c for given probabilities, Table A8

~ X be DOnnIJ with mean .5 and variance 0.04 (beoce s1and8Jd deviation 0.2). PiIMI c (W k ~ponding 10

the given probability

P(XSc) -~.

P(-s - t s X ~ S + t) z -.,

1'(X ~ c) - I~.

EXAMPLE 4 Defectives

In a production of iron rods let dte diameter X be normally distritMlted with mean 2 in. and standard deviation

0.008 in.(a) What percentage of defectives can we expect if we set the tolerance limits at 2 ~ 0.02 in.?

(b) How should we set dte tolerance limits to allow f(X" 4% defectives?

Solution. (a) 11% because from (5) and Table A7 we obtain for dte complementary event dte probability

p(1.98 ~ x S 2.02) - . (2.02 - 2.00) - . (1.98 - 2.00)0.0 0.-

(b) 2 ~ 0.0164 bccause f« die complement.:y event we have

0.96 = P(2 - c ~ X ~ 2 + c) «

so !hat Table AS gives

(2+C-2 ) 2+c-2 0.98 = III 0.008' 0.008 = 2.054. c = 0.0164. ~~ .':...':. I') "...

Binomial Distribution Approximated by Normal Distribution

The probability function of the binomial distribution is (Sec. 22.7)

(8) f(x) = (:) pzqn-z (x = 0, I, . . . , n).

... . If n is large. the binomial coefficients and powers become very inconvenient. It is of great

practical (and theoretical) imponance that in this case the nonnal distribution provides agood approximation of the binomial distribution. according to the following theorem. oneof the most important theorems in all probability theory.

1089Normal Distribution

(X - 0.8 2.44 - 0._)P(X S 2.44) - ,. 2 S 2 - 1'(% S 0.82) - 0.7939

( 1-0.8 )P(X ~ I) = I - 1'(X ~ I) - I -. --;:- - I - 0.5391 - O.~

...p(1.0 ~ x ~ 1.8) - ~O.S> - ~O.I) = 0.691' - 0.5398 - 0.1'17.

(c - ") c - .5. 0:2 -~. -o:r- - 1.64.S.C R 5.329

'+1-5.329 (as befcwe; why?)

c-5~ - 2.326. c - 5.4652. ~1bUI p(x a c) - ~.

s 4I(2.~) - ~-2.'> - 0."38 - (I - 0."38) = 0.9876 - 98"'.

0.98 - P(X~ 2 + c)

(~) rq,,-a-

1090 Data Analysis. Probability Theory - Chap. 22-

THEOREM 3 (Limit theorem of De Moivre and Laplace)

For large n.

(9) f(x) - f.(x) (x = 0, 1, . . . ,II).

Here f is givell by (8). The fwaction

(10)

is the density of the normal distribution with mean IJ. = np and variance u2 = npq (themean and variance of the binomial distribution). The symbol - (read asymptoticallyequal) means that the ratio of both sides approaches 1 as n approaches 00. Furthennore.for any nonnegative integers a and b (> a),

(11)

A proof of this theorem can be found in [G3] listed in Appendix 1. The proof showsthat the term 0.5 in a and ~ is a correction caused by the change from a discrete to acontinuous distribution.

1. Let X be normal with mean 10 and variance 4. Find P(X > 12). P(X < 10), P(X < II),P(9 < X< 13).

1. Let X be nOIma1 with mean 105 and variance 25. Find P(X ~ 112.5). P(X > 100),,,:. . P( II 0.5 < X < 111.25).-.." ,,.r.I'... 3. Let X be nonnal with ~ ~ and variance 9. Detennine c such that P(X < c) = 5111.

.' P(X> c) = 1111. P(~ - c < X < 50 + c) = 50%.

4. Let X be nonnal with mean 3.6 and variance 0.01. Find c such that P(X ~ c) = 50'10.P(X > c) = 10%. P(-c < X - 3.6 ~ c) = 99.9111.

5. If the lifetime X of a certain kind of autonK>bile battery is normally distributed with a mean of5 years and a standard deviation of I year. and the manufacturer wishes to guarantee the batteryfor 4 years. what ~tage of the batteries will he have to replace under the guarantee?

. 6. If the standard deviation in Prob. 5 were smaller. would that percentage be larger or smaller?7. If the resistance X of certain wires in electrical networks is normal with mean 0.01 ohm and

standard deviation 0.001 ohm. how many of 1000 wires will meet the specification that theyhave resistance between 0.009 and 0.011 ohm?

8. What is the probability of obtaining at least 2048 heads if a coin is tossed 4040 times and headsand tails are equally likely? (See Table 22.1 in Sec. 22.3.)

51

1 -../2~ ,r(.I') - '\I2-;v';;;;q X-lip

z=-v~

b~ (:) p.q"-% - ca.(.8) - <I»(a),

%-4P(a ~ x ~ b) =

~ = b - lip + 0.'

v;;pq

a = a - lip - 0.5v;;;;q .

Sec. 22.9 Distributions of Several Random Variables

9. If the mathematics scores of the SAT college entrance exams are nonnal with mean 480 andstandard deviation 100 (these are about the actual values over the past years) and if some collegesets 500 as the minimum score for new students, what percent of students would not reach thatscore?

10. A producer sells electric bulbs in cartons of 1 <XX> bulbs. Using (II). find the probability thatany given carton contains not more than I % defective bulbs. assuming the production processto be a Bernoulli experiment with p = 1% (= probability that any given bulb will be defective).First guess. Then calculate.

II. If d1e monthly machine ~r and maintenance cost X in a certain factory is known to be normalwith ~ $12<XX> and stalMiard deviation $2<XX>, what is d1e probability that the repair cost forthe next month will exceed the budgeted amount of $1 S<XX>?

11. The breaking strength X [kg] of a certain type of plastic block is nonnally distributed with amean of 1 SOO kg and a standard deviation of 50 kg. What is the maximum load such that wecan expect no m<xe than 5% of d1e blocks to ~?

13. If sick-leave rime X used by employees of a company in one month is (very roughly) nonnalwith mean 1000 hours and standard deviation 100 hours, how much time t should be budgetedfor sick leavc: during the next month if t is to be exceeded with probability of only 20%?

14. TEAM PROJECT. Normal Distribution. (8) Derive the fonnulas in (6) and (7) from theappropriate normal table.

(b) Show that <II( - z) = I - <II(z). Give an example.

(c) Find the points of inflection of the curve of (I).(d) Considering <112(~) and introducing polar coordinates in the double integral (a standard trick

worth remembering). prove

(12)

(e) Show that u in (I) is indeed the standard deviation of the nonnal distribution. [Use (12).J(f) Bernoulli's law of large numbers. In an experiment let an event A have probability p

(0 < p < I), and let X be the number of times A happens in n independent trials. Showthat for any given f > O.

(g) Transfonnaoon. If X is nonnal with mean Jl. and variance u2, show thatX. = cIX + C2 (Ct > 0) is nonnal with mean Jl.. = CtJl. + C2 and variance 0'.2 = c.2u2.

WRITING PROJECT. Use or Tables. Give a systematic discussion of the use of Tables A 7and A8 for obtaining P(X < b), P(X > a). P(a < X < b). P(X < c) = k. P(X > c) = k.P(Jl. - c < X < Jl. + c) = k; include simple examples.

15.

-:. . ,..~;,.. "~ '-0.

22.9 Distributions of Several Random VariablesDistributions of two or more random variables are of interest for two reasons:

1. They occur in experiments in which we observe several random variables, forexample, carbon content X and hardness Y of steel. amount of fertilizer X and yield ofcorn Y. height Xl. weight X2. and blood pressure Xa of persons. and so on.

2. They are needed in the mathematical justification of the methods of statistics inChap. 23.

1091

1 .~(~) - ~ f ~-.-ndJI-

v2. -. I.

"(1; - pi s .) - 1 as n - ~.