Quiz #2 of Probability and Statistics (I) Date: December 9, 2005
1. The average service speed in the check-in queue of hotel is 0.2 customer per minute. If you are now in
2.
3.
4.
5.
6.
the third position of the waiting line, please find the probability of finishing the check-in procedure for
you within 5 minutes. (10%)
A basketball player makes 90% of her free throws. What is the probability she will miss for the first time
on the seventh shot? (10%)
In a certain city, the daily consumption of electric power (in millions of kilowatt hours) is a random
1 x3
 xe for x  0
variable having the probability density function, f ( x)   9
. If the city’s power plant has
0
for x  0
a daily capacity of 12 million kilowatt-hours, what is the probability that this power supply will be
inadequate on any given day? (10%)
The time for a super glue to set can be treated as a random variable having a normal distribution with
mean 30 seconds. Find its standard deviation if the probability is 0.20 that it will take on a value greater
than 39.2 seconds. (10%)
From experience a famous construction firm found that the lowest bid on a construction job can be
2C
 3
for
 x  2C

regarded as a random variable X having the uniform density function, f ( x)   4C
,
3
0
elsewhere
where C is his own estimate of the cost of the job. What percentage should the manager of this firm add
to his estimated cost when submitting the price bidding in order to maximize the job expected profit?
(10%)
Given that the switchboard of a consultant’s office receives on the average 0.6 calls per minute, find the
probability that the time between successive calls arriving at the office will be (a) within one minute.
(10%) (b)What differences will be if you apply the Poisson distribution in contrast to the Exponential
one for the (a) question? (10%)
 x x for 0  x1  1, 0  x2  2
7. If two random variables have the joint density function, f ( x1 , x2 )   1 2
,
elsewhere
0
find the probabilities that the sum of the values taken on by the two random variables will be less than 1.
(10%)
6
2
 ( x  y ) for 0  x  1, 0  y  1
8. If two random variables have the joint density function, f ( x, y )   5
,
0
elsewhere
find the expressions for the following probability density functions, (a) f(x/y) (10%) and (b) f(x/y=0.5).
(10%)
Note:
1.Poisson
distribution:
P(x;λ)=
x e 
x!
for x  0,1,2,...,   0 ;
Exponential
distribution:
 1  x
for x  0,   0
 e
P(x;β)=  
, (Supposed e=2.7183)
0
elsewhere

2. Some referral values extracted from the cumulative probability tables may be useful to you.
F(x,λ) is the cumulative distribution function of Poisson. F(1,1)=0.736,F(2,1)=0.920,F(3,1)=0.981,
F(4,1)=0.996,F(5,1)=0.999,F(0,0.2)=0.819,F(1,0.2)=0.982,F(2,0.2)=0.999. Z(x) is the cumulative
distribution function of standard normal. Z(0.2)=0.5973,Z(0.84)=0.7995,Z(0.85)=0.8023
Solutions:
1. Let X be a Poisson distribution with λ=1. You will be served within 5 minutes if x≧3. P(x≧3)=1-P(x≦
2)=1-0.920=0.080
2.
It is a geometric distribution. g(x;p)=g(7,0.1)=(0.1)(0.9)6=0.048
set u  x, e
3.
x
3
dx  dv, du  dx,3e
x
3
 1 
1 x3
1
x
x
xe dx  (3xe 3 )   (3e 3 )dx
12 9
12 9 12
9
 v, 


1
x
 4e 4  (9e 3 )  4e 4  e 4  5e 4  0.0916
12
9
4.
X is a random variable of normal distribution. ∴Prob(X>39.2)=Prob[(X-30)/σ>(39.2-30)/ σ]=0.2,
(9.2/σ)=0.842 (by the interpolation method), σ=9.2/0.842=10.93
5.
Let’s set the manager adding k percentage to cost as the bidding price, (1+k)C, then the profit will be
kC if he wins the bid. The expected profit under competition:
2C
2C
2C
3
3
3k
(1k )C kC( 4C )dx  k (1k )C 4dx  4 ( x) (1  k )C  (3k 4)(1  k )C, F .O.C., 3k 2  3 4  0, k  0.5 ,∴ the
6.
added percentage on cost should be 50%
By using the exponential distribution, β=1/λ=1/0.6=5/3 (minute),∴
Prob(t  1)  (e
3 x
1
)  1  e 0.6  0.451
0
5
0  0.6
by Poisson, P(x  1,   0.6)  1 - P(x  0,   0.6)  1 - (0.6) e
1 1 x2
Prob( x1  x 2  1)  

0 0
7.
 1  e 0.6  0.451, the smae result as (b)
1
1
2 1  x2
x 2 x1 dx1 dx 2   x 2 ( 1 x1
)dx 2  1  x 2 (1  x 2 ) 2 dx 2
2
2
0
0
0
2
3
41
 1 ( 1 x2  2 x2  1 x2 )  1  1  1
2 2
3
4
2 12
24
0
(a) f ( x / y ) 
8.
0!
16
1
1
f ( x, y )
, f ( y )   ( x  y 2 )dx  6 ( 1 x 2  y 2 )  6 ( 1  y 2 ),
5 2 0
5 2
0 5
0
f ( y)
6 (x  y 2 )
x  y2
 f ( x / y)  5

, for 0  x  1 and 0  y  1 0, 0 for elsewhere
6 (1  y2 ) 1  y2
2
5 2
x  0.25
4x 1
(b) f ( x / 0.5) 

 , for 0  x  1, 0, for elsewhere
0.5  0.25 3 3