Download Probability Theory Homework #5

STA 131A Probability Theory
Fall 2011
Homework #5
1.3 If X is the r.v. denoting the number of tickets to win a prize, then X ,...,
B(50,0.01). Therefore:
(i) P(X
= 1) = (1'1°)(0.01)(0.99)49:: 0.306.
~ 1) = 1 - P(X = 0) = 1 - e~~)(0.01)0(0.99)50 :: 1 _ 0.605 =
(ii) P(X
0.395. •
1.6 If X is the r.v. denoting the number of time the bull's eye is hit, then
X '" 8(100, pl. Therefore:
= E~'.!?40 (J~)p%ql00-% (q = 1- pl.
(ii) P(X ~ 40) = E~'.!?40 e~)(0.25)%(0.75)IOO-z.
(iii) EX = np = 1001' and Var(X) = npq = 100pq, and for p = 0.25,
we have: EX = 25, Var(X) = 18.75, and !:I.d. of X = v'18.75 ::
(i) P(X ~ 40)
4.33. •
1.7 Here EX
= np = I~ = 25 and
P(IX - 251
< 10) Var(X) = npq
=
P(IX - EXI
=
1 - 0.1875
= 25 x i
= 18.75, so that:
u 2 (X) < 10) ~ 1 - ~
= 0.8125.
=1-
18.75
100
•
=
1.12 Here (n + 1)1'
J~J = 25.25 is not an integer. Then there is a unique
mode obtained for x = 25. In other words, f(25) is the unique maximum
probability among the probabilities f(x)
P(X = x), x = 0,1, ... , 100,
and one would bet on the value X = 25. Also, f(25) ::::: 0.092. •
=
1.14 Here X has the Geometric distribution with I'
(0.01)(0.99)Z-J,x = 1,2, ... Then:
P(X ~ 10) = 0.01,
=
1 - P(X ~ 11) = 1 - (0.01)(0.99)1°11
=
1 - (0.01){0.99)10
X
1 _ ~.99
=1-
= 5) = e- 2•3 (2.~)S :: 0.054.
5.
=
+ 0.99 + (0.99)2 + ...J
(0.99)10:: 0.096.
1.20 Since P(X = x) = e-.l.~,x = 0, 1, ... , we have: P(X
so that -.\ = log(O.I) = -2.3 and .\ = 2.3. Then:
P(X
so that f(x)
•
= 0) = e-.I. = 0.1,
•
1.24 (i) An appropriate probability model here is the Poisson distribution
with parameter 3, P(3).
{ii) What we are asked to find here is the mode (or modes) ofthe distri­
bution. Since .\ = 3, Theorem 2 provides for two modes 3 and 2. So,
the number of items which arrive within 1 second with the maximum
probability is either 2 or' 3. The respective probability is 0.2241. •
1.26 We have:
1.27 (i) The distribution of X is Hypergeometric with m
110 that:
= 15 and n = 10,
x = 5, ... ,IS.
(iii) This probability is 0, since there are only 10 specimens from the rock
R2' •
1.30 We have P(X = x) =_, (:)(0.15)&'(0.85)n-&'. x = 0.1, ... , n, 110 that
P(X = 0) = (0.85)n, P(X = 1) = n(O.15)(0.85)"'-I. Therefore,n(O.15)(0.85)",-1 >
(0.85)n, or 0.15n > 0.85. or n >
= ~ ~ 5.667, 110 that n = 6. •
t.ft
1.32 Let X be the r. v. denoting the number of defective items among the
100 sampled. Then the r.v. X has the Hypergeometric distribution with:
m = 3, n = 997, r = 100. Therefore:
(i) The probability of accepting the entire shipment is:
P(X ~ 1)
P(X
=::
=0) + P(X =
1)
= (~)(i:)
(\':)
+
m(:')
(~~)
3 x (:') _ 538,501 _
I ) + (1000) - 553 890 - 0.97.
=
~
100
100
= t~ = 0.3.
(iii) Var(X) = 3x997xl00(I,OOO-I00) =
(t,OOO)'(I,OOO-I)
el:»
'
(ii) EX
vy·~:/ /10 ~ 0.519.
2,991
l00x 111
~
0.269, and O'(X)
=
•
1.35 Let X be the r.v. denoting the number of homeowners who have earth­
quake insurance. Then X has the Hypergeometric distribution with m =
125, n = 875, r = 40. Hence:
(i) EX
= ...m!....
125
mnrfm+n-r)
=
m+n = ~
1,000 = 3.
, 0'2(X) = (m+n
i(m+n-l) = 126x875x26x975
1,0002 x999
Ila~bV: ~ 2.669, and O'(X) ~ 1.634.
(ii) P(3.125 - 1.634 :5 X :$ 3.125 + 1.634) = P(1.491 :5 X :5 4.759)
P(2:5 X ~ 4) = ~ E!=2 e~6)(2~~2:)' denote it by Q.
~
=
21 }
(iii) Since m~n = 1~:O = 0.125 = 2/16, the Binomial approximation (see
Exercise 1.29 in this chapter) gives a ~ E!=2 (:S)(fe)&'(~)26-&' =
0.8047 - 0.1623 = 0.6424. Next, 25 x 0.125 = 3.125 and the Poisson
Tables give: Q ~ 0.8153 - 0.1991 = 0.6162 for A = 3.00, and a ~
0.7254 - 0.1359 = 0.5895 for A = 3.50. By linear interpolation, we
get Q ~ 0.6095. •