Homework 4 (100pt) Due in class Friday 9/21

Homework 4 (100pt) Due in class Friday 9/21
1. (8) 1.59 (page55)
X~Binomail(1000, 1/200)
Pr ( X  8)  1  Pr ( X  7)  1  0.8672  0.133
Grading note: some might use Poisson approximation and get an answer of 0.131
Pr (5  X  10)  Pr ( X  5,6,7,8,9,10)  0.546
Grading note: some might use Poisson approximation and get an answer of 0.544
2. (8) 1.67 (page57)
(a)
The area under the density must equal 1. To verify this, make the substitution u = x2/(22),
which gives du = xdx/2 , so


0
x

2
e x
2
/(2 2 )
dx =


e u du =  e u
0
upper limit x = c, the integral becomes: Proportion(x  c) =

c
0
1 - ec
2
/(2 2 )


0
x
2
= 0 - (-e0) = 1. For a finite
e x
2
/(2 2 )
dx =  e u

c 2 /(2 2 )
0
.
(b) For  = 100, Proportion(x  200) = 1 - e 200 /( 2(100)
2
2
)
= 1 - e-2 = 1 - .135335 = .8647.
Proportion(x  200) = 1 - Proportion(x < 200) = 1 - .8647 = .1353.
Proportion(100  x  200) = Proportion(x  200) - Proportion(x  100) = .8647 - [1-
e 100 /( 2(100) ) ] = .8647 - [1-e-.5 ] = .4712.
2
2
3. (8) 1.73 (page 58)
Pr(a randomly selected bottle has a bursting strength > 300 psi)=Pr(X>300)=Pr(Z>1.67)=10.9525=0.0475
Define Y as the number of bottle in a carton that have a bursting strength >300 , Y~Binomial(12,
0.0475)
12 
0
12
 0.0475 (1  0.0475) =1-0.5577=0.4423
0
Pr(Y  1)= 1- Pr(Y=0)=1- 
=
4. (16) Case study. Determine the PMF for the following random variables. Identify which one(s) of the
random variable is Binomial.
a) Of all customers purchasing automatic garage-door openers, 75% purchase a chain-driven model.
Let X be the number among the next 15 purchaser who select the chain-driven model
Define the PMF of X and decide whether X is Binomially distributed.
15 
x
15 x
 0.75 (1  0.75)
x
X ~ Binomial(15, 0.75) , the PMF of X: Pr(X=x)= 
b) A pediatrician wishes to recruit 5 couples, each of whom is expecting their first child, to
participate in a new natural childbirth regimen. Let p  0.2 be the probability that a randomly
selected couple agrees to participate. Define X be the number of patients the pediatrician asks
before she finds 5 couples who agree to participate. (Hint: consider the case where she needs
only 1 couple, what is the probability that she asks 1 couple to find 1 couple who agree, how
about the probability that she ask 2 couples to find 1 couple? then consider the case where she
needs 2 couples, what is the probability that she asks 2 couples to find 2 agree, how about asks 3
couples to find 2 agree… until you find the pattern)
X is not Binomial because sample size n is not fixed.
 x  1
x 5
5
 (1  0.2) (0.2)
5

1


The PMF of X: Pr(X=x)= 
This is because for a total of x couples she asks, 5 couples agree , hence (0.2)5 , x  5 couples
do not agree , hence (1  0.2) x5 . Also for a total of x couples, the last one must agree, and the other 4
 x  1

 5 1
who agree can be arranged in the first x  1 , hence 
Define the PMF of X and decide whether X is Binomially distributed.
c) Each of 12 refrigerators of a certain type has been returned to a distributor because of an audible,
high-pitched, oscillating noise when the refrigerators are running. Suppose that 7 of these
refrigerators have a defective compressor and the other 5 have less serious problems. If the
refrigerators are examined in random order, let X be the number among the first 6 examined that
have a defective compressor. (Hint: for example, the total number of ways to arrange 7 defective
 12 
 . Suppose there are 2 defective happen in the first 6 examined,
7
refrigerators out of 12 is 
6
 2
these 2 can happen in a total of   ways; then the second 6 must contain 7-2=5 defectives,
6
5
these 5 defectives can happen in a total number of   arrangements, what is the probability
that 2 defectives are arranged to happen in the first 6 examined? )
Define the PMF of X and decide whether X is Binomially distributed.
X is not Binomial because p is not fixed.
 6 6 
 

x7  x

The PMF of X is Pr(X=x)=
, where x  1, 2,3, 4,5,6
12 
 
7
5. （10） 1. 30 (page 41)
(a) Proportion(z  2.15) = .9842 (Table I). Proportion(z < 2.15) will also equal .9842 because
the z distribution is continuous.
(b) Using the symmetry of the z density, Proportion(z >1.50) = Proportion(z < -1.50) = .0668.
Proportion(z > -2.00) = 1- Proportion(z  -2.00) = 1 - .0228 = .9772.
(c) Proportion(-1.23  z  2.85) = Proportion(z  2.85) - Proportion(z  -1.23) = .9978 - .1093
=.8885.
(d) In Table I, z values range from -3.8 to +3.8. For z < -3.8, the left tail areas (i.e., table values)
are .0000 (to 4 decimal places); for z > +3.8, left tail areas equal 1.0000 (to 4 places).
Therefore, Proportion(z > 5) = 1 - Proportion(z  5) - 1 - 1.0000 = .0000. Similarly,
Proportion(z >-5) = 1 - Proportion(z  -5) = 1 - .0000 = 1.0000.
(e) Proportion(z < 2.50) = Proportion( -2.50 < z < 2.50)
= Proportion(z < 2.50) - Proportion(z < -2.50) = .9938 - .0062 = .9876.
6. （10） 1.32 (page 41)
(a)
Proportion(z  z*) = .9082 when z* = 1.33 (Table I).
(b) Proportion(z  1.33) = .9082 and Proportion(z  1.32) = .9066; using linear
interpolation, .9080 is (.9080-.9066)/(.9082-.9066) = .0014/.0016 = .875 (or, 87.5%) of the
way from .9066 to .9082, so z* is .875 of the way from 1.32 to 1.33; i.e., z* = 1.32
+ .875(1.33-1.32) = 1.32875, or, about 1.329.
(c) Proportion(z > z*) = .1210 is a right-tail area. Converting to a left-tail area (so that we can
use table I), Proportion(z  z*) = 1 - .1210 = .8790. From Table I, z* = 1.17 has a left tail area
of .8790 and, therefore, has a right-tail area of .1210.
(b) Proportion(-z*  z  z*) = .754. Because the z density is symmetric the two tail areas
associated with z < - z* and z > z* must be equal and they also account for all the remaining
area under the z density curve. That is, 2Proportion(z  - z*) = 1 - .754 = .246, which means
that Proportion(z  - z*) = .1230. From Table I, Proportion(z  -1.16) = .1230, so - z* = -1.16
and z* = 1.16.
(c) Proportion(z > z*) = .002 is equivalent to saying that Proportion(z  z*) = .998. From Table I,
Proportion(z  2.88) = .9980, so z* = 2.88. Similarly, you would have to go a distance of 2.88 (i.e., 2.88 units to the left of 0) to capture a left-tail area of .002.
7. （10） 1.38 (page 42)
(a)
When standardized, x > 100 becomes z > (100-96)/14 or, z > .29. From Table I,
Proportion(z > .29) = 1 - Proportion(z  .29) = 1 - .6141 = .3859.
(b) When standardized, x  75 becomes z  (75-96)/14 or, z > -1.5. From Table I, Proportion( z  -1.5)
= 1 - Proportion(z < -1.5) = 1 - .0668 = .9332.
(b) When standardized, 50 < x < 75 becomes (50-96)/14 < z < (75-96)/14 or, -3.29 < z < -1.5.
From Table I, Proportion(-3.29 < z < -1.5) = Proportion(z < -1.5) - Proportion(z < -3.29)
= .0668 - .0005 = .0663.
(c) Proportion(x  a) = .05, when standardized, becomes Proportion(z 
a 96
14
) = .05. From Table
I, 5% of the area under the z curve lies to the left of z = -1.645. Therefore,
a = 96 -1.645(14) = 72.97. For the right-tail area,
119.03.
b 96
14
a 96
14
= -1.645, so
= +1.645, so b = 96+1.645(14) =
8. (12) Suppose that the reaction time (sec) to a certain stimulus is a continuous random variable with a
density function given by
f ( x)  {
k / x 1  x  10
0
otherwise
a) Find the numerical value of k

10
1
k / x  1  k= 1/ln10=0.434
b) Find the mean and std of x
10
E ( X )   xk / xdx  k (10  1)  9k  9 / ln10 =3.906 or 3.91
1
 X  E ( X 2 )  ( EX )2 = 49.5k  (9k )2 = 6.226 =2.495 or 2.5
c) Find the mean of 2x
E(2 X )  2*3.906  7.812
d) Find the mean of x 2 .
10
E ( X 2 )   x 2 k / xdx = 49.5k  21.483
1
9. (8) Use normal approximation on Binomial to compute the following questions with continuity
correction , suppose X ~ Binomial (500,0.15)
a). Pr ( X  400)
X ~ Normal (500(0.15), 500(0.15)(1  0.15)) or X ~ Normal (75,7.984)
Pr ( X  400)  Pr ( X  399)  Pr ( X  399.5)  1
b). Pr (200  X  450)
Pr (200  X  450)  Pr (200.5  X  450.5)  0
10. （10） Discussion question. Discuss on-line via Blackboard Learn. Discuss only, no need to turn
in your work, grading will be based on your level of participation.
a). Suppose X is the cost of an appetizer and Y be the cost of a main course at a certain restaurant .
Suppose a customer always order both courses. The price and probability of an order is given below:
Note Pr ( X  5, Y  10)  0.2 means the probability of order a $5 appetizer AND a $10 main
course is 20%.
Pr ( X
Pr ( X
Pr ( X
Pr ( X
Pr ( X
Pr ( X
Pr ( X
Pr ( X
Pr ( X
 5, Y  10)  0.2
 5, Y  15)  0.15
 5, Y  20)  0.05
 6, Y  10)  0.1
 6, Y  15)  0.15
 6, Y  20)  0.1
 7, Y  10)  0.1
 7, Y  15)  0.1
 7, Y  20)  0.05
Do you think the order (cost) of appetizer and main course are independent?
b). Suppose X ~ Normal ( ,  ) ， Y ~ Normal ( ,  ) ，
What is E(X+X), Var(X+X), E(X+Y) and Var(X+Y).