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Unformatted text preview: 514 12 Probability and Statistics Case 2: (P(X = 1) P(Y = 1))/(P(X + Y) = 2) == 0.022308770/0044617539 = 0.5000 Case 3: is identical to case 1. Notice that the probabilities of the above three cases add up to
exactly unity. Prob. 81. Since only 1 in 1000 experiences side effect p = 1/ 1000 Since 500 is the strength of the
test group, n = 500. We shall use Poisson approximation approach. it = n p = 500 I! 1000 = 0.5
P(X = k) = if" .ik/kl P(X = 0) = 060653 is the probability that none of the 500 persons
experiences side effect Prob. 83. (a) The probability that there are no cases of Down’s syndrome P (X = 0) = (699/700) 1039. ; 0.2394 The probability that there is exactly one case of Down’s syndrome P0) = C0000, I) 
(699/700)”9  (1/200) = (1000)  (0.239748)  (1/700) = 0.342498. Hence the probability
that there is at most one case is: P(O) + P(l) = 0.5818 (b) We are using a Poisson approximation in = n  p = 1000 ' 1/700 = 1.42857 P(X = k) =
e—JL  21" /k! P(X = 0) = 0.2396 P(X = 1) = 0.3423. Hence the probability that there is at
most one case is P(X = 0) + P(X = 1) = 0.5819 Prob. 85. There are P = 5.0 parasitoids, vying with one another to ﬁnd a host. The probability
of any one of them encountering a host is a = 0.03 Hence, the probability that the host escapes. detection is (for such cases, we take the only the ﬁrst term of Poisson distribution ) is 3—9350 =
e‘” = 022313016 12.5 Continuous Distributions Prob. 1. As per Equation 12.28 in Example 1 in Section 12.5.1, any non~negative function which
satisﬁes the condition 00
f f (x)dx z 1
—eo
is a candidate for being a density function. Carrying out the indicated nonnegative test, 36—30: 00 =1
_3 0 00 00
f (swoon: = f (3e"3")dx =
—co 0 Thus f (x) is indeed a probability density function. The corresponding distribution functi:
be seen, from above, to be: —e“3x forx 2 0 F(x)={o forx<0 Note that F (x) is discontinuous at x = 0. Discontinuities are permitted for P(x) 'buf'not._ aim Prob. 3. (
density fir
Let x = ta Hence, we Prob. 5. It is requii The variai Hence the Prob. 7. The densi EX canb 12.5 Continuous Distributions 515 Prob. 3. Given that f (x) = c/ (1 + x2), it is necessary to determine 0 so that f (x) is a probability
density function. All we need to do is to ﬁnd the area under the curve for overa all real numbers.
Letx = tune; then dx = sec29d9 [0° dx [#2 ( 2 0) are
C = C SEC —'.""—" = CH
00 1 i x2 _,,;2 (1 + tanzn') 1
Hence, we conclude that c = —
3T Prob. 5.
2e‘2" forx > 0 Given density function f (x) = { 0 font < 0 It is required to ﬁnd Expectation and Variance. 00 {x}
EX = f (x23'2‘dx = xerl‘ + (0.5%?”
D 0 The variance is found from the usual expression (E X 2 — (E102) be 00 00 1
EX2 = / (x2)2e_2")dx = —x28_2x + f 2xe‘23dx = —
o o o 2
I l 1
th '. ' — — — 2 = —
Hence e variance 1s 2 (2) 4 I Prob. 7. In this problem we are given the distribution function F (x) to begin with.
F(x>=(1—1/x2) » The density function is hence:
2
f0?) = —3
.x
EX can be found by a simple integration, to be 00. (—2/x) = —2 1 E X 2 can also be found, by another simple integration to be —21n(x_). Using the given lower and upper limits, we write:
00 —_2 ln(x) The interesting point is that at the upper limit, the value of 1n(x) is unbounded. Hence Var(x) is
unbounded. 516 12 Probability and Statistics 12.5 Cont Prob. 9. it pl (a — Dar—'5' forx > 1 Given density function f (x) = { 0 folx < 1 EX2 ll 00
f (e — 1)x1‘“dx
l «a — I)/(2 — aux“
l
unbounded if a < 2 ,_:_ 00 ll (a) If a < 2 then EX is unbounded, since 33‘“ increases without limit. Not:
(1)) shat
EX=(a—1)/(a—2) if a>2 .
(d) Grai
Prob. 11. (a) To prove that the curve is symmetric about 3: = ,u., we need to show that f (x +
2
p.) = f(x — u). Deﬁne K = ”(ax/2n, f(x) = Kira—“Fa” , then we need to show that
f(# +x) = f (.u. — x). Obviously vie—(“+1”): = .¢3‘“U‘"J“""‘}2 thus, it is clear that the normal
distribution curve is symmetric about it = y.
(b) To'prove that the maximum of the curve is at x = a, we need to show that the ﬁrst derivative
is zero at that point. Let us also make the substitution 1: = 02
1
Let f(x) = [WW/2", we only need to prove that df (x) {dx = o and f'(x) = —;(x 
u)f(x), it is clear that at x = u. f’(x) = 0.
Note that we have omitted K in the above expression, as it plays no role in differentiation(6t
in the conclusion) Prob. 13_
(c) Now, the problem is to establish the inﬂection points. At such pointsthe curve does not we can ev: go through a maximum or a minimum though the derivative goes through zero. The curve
goes upward, stops for a while as if to decrease and then changes its mind and continues:
its journey upwards! That is why the ﬁrst derivative goes through zero, though the curve
does not have a maximum at that point. So, we need to examine the second derivative at From the!
such a point. In order to reduce clutter in long algebraic expressions, let us ﬁrst simplify; the
distribution function by making appropriate substitutions. Let us also omit the constant Kas 12.5 Continuous Distributions 517 it plays no role here. Nonnal Curve f(x) = Ke(x—Wzi’zaz
Put (x — ,u) = 2:
Then dx = dz
Variance v = 02
K can be omitted for simplicity
Incorporating above changes f (z‘) = e'zzﬂ"
Differentiating once f’Cz) = (—Z/ 10f (Z) (1/ WW” ’(z) + f (z)
(rm/m1 — zz/v) 0 at z = in p. i a Differentiating again f " (z)
Simplifying f"’(z) It is clear that f ”(2) Thus, inﬂection ocean at x Note that f(z) is obtainable from ﬁx) by a mere translation by ,u on the x axis. Hence the
shape of the curve is preserved (d) Graph of f(x_) for p, = 2 and a = 1: Prob. 13. Given that p. =12.8 and 0' = 2.7 we can evaluate the following:
,u — 20' = 7.4 and. p. + 20 = 18.2 From the graph of normal distribution, it is known that 95 percent of the area falls between p—Za anda+20 518 12 Probability and Statistics i.e. between 7.4 and 13.2 respectively. Hence
P(X e [7.4, 18.2] = 0.95) In a similar manner, it is possible to ﬁnd an interval such that 99 percent falls within this interval.
From the graph of normal distribution it is known that 99 percent of the area falls between ,u.—30' and gt+3cr This works out to 4.? and 20.9 respectively. Hence P(X E [4.7, 20.9] = 0.99). Prob. 15. In this problem, we are asked to determine the fraction of population that will fall in the interval P(,u g X c: 00). Since the normal distribution curve is symmetric about 1.1., the area to l
the right of ,u. is onehalf the total area. Hence the answer to the question is 2’ Prob. 17. Therange mentioned in this problem, which is similar to the previous one is: ~00 and
(u + 30). p’ 1
Area = — Known fact (12.1)
00 2
ﬂ+3a
Area = 0.995 Known from Appendix B (12.2)
31,—30
“+30
Hence Area = (0.995/2) = 0.495 (12.3).
it
,u.+30r
Adding (1) and (3) above Area = (0.5 + 0.495) = 0.995 (12.4)
~00
(12.5) Hence the fraction of population that will fall in the range given in this problem is 0.995 Prob. 19. In this problem, both the lower and upper limits of X are to the left of p. The range given
is (—00) to a — 20) . The fraction of population in this range is: (1.0 — 0.95)/2 = 0.025 Prob.21. Foru=3andar :2 (a) The probability P(X s '4) is to be determined. The normalized random variable Z .=
(X — ,u)/a is (4 ._ 3) /2 = 0.5. From Appendix B, the area under the standard normalised
curve to the left of Z = 0.5 is 0.6915. Hence P(X 5 4) = 0.6915. In other'words. were
is a 69.15 percent probability that the random variable is less than or equal to 4. Since parts
(b), (c), (c) of this Problem are similar to (a) let us consider tabulating the key results. 12.5 Conti (b) In 01
perm (c) Thor
abov To 13'
is 15 (d) In th
in A} ...
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 Winter '07
 SCHONMANN

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