#5 Due 10/8/041. The total number of hours measured in units of 100 hours that a family runs avacuum cleaner over a period of one year is a continuous random variable withprobability density function

f ( x )=x 0 < x < 1

2 − x 1< x < 2

0 elsewhere

A. What is the cumulative distribution function of X?

For 0 < x < 1

F(x) = tdt =x2

20

x

∫F o r 1 <x < 2

F(x) = tdt0

1

∫ + (2 − t)dt1

x

∫ =12

+ (2t −t2

2)

1

x

=12

+ x −x2

2−

32

Thus

F ( x )=

0 x < 0x2

20 < x < 1

−1+ x −x2

21< x < 2

1 x > 2

B. What is the probability that over the course of a year a family runs the vacuumcleaner

(1) (2) less than 60 hours?

F(.6) =

(.6)2

2= .18

(3) between 60 hours and 120 hours?

F(1.2) − F(.6) = −1+1.2 −

(1.2)2

2− .18 = .3

C.Find the mean, median and the variance of random variable X?

EX = x2

0

1

∫ dx + x(2− x)dx=11

2

∫

EX2 = x3

0

1

∫ dx + x2 (2 − x)dx=761

2

∫σ2 =

16

Median =1 by symmetry

2. The waiting time x in hours between successive speeders spotted by a radartrap is a continuous random variable with cumulative distribution

F ( x )=

0 x ≤ 0

1− e−8x x > 0

A.What is the probability density function?

f ( x )= 8e−8x , x> 0

B. What is the probability that(1) the waiting time between speeders is less than 12 minutes?

F(.2) =1− e−8(.2) = .798

(2) the waiting time between speeders is at least 10 minutes.

1− F( 16 ) = e−8 / 6= .264

C. Find the mean waiting time between speeders.

EX = 8xe−8x

0

∞

∫ dx = limA→ ∞

−xe−8x

0

A+ e−8xdx

0

A

∫

= limA→ ∞

−xe−8x

0

A−

18

e−8x

0

A

=18

hr

u = x

du = dx

dv = 8e−8xdx

v = −e−8x