Download The distance between major cracks in a highway follows an

Name:______Answer
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1.
5
2.
3.4657
3.
0.0498
4.
(b)
5.
(b)
6.
(c)
7.
40320
8. 105  / 16 =11.6325
Quiz 4 (2/23/04 in class)
The distance between major cracks in a highway follows an exponential distribution with a mean of 5
miles. Answer the following 3 questions using this information.
1.
What is the standard deviation of the distance between major cracks?
Notice that E(X)=
2.
Var(X ) =1/=5 miles in exponential distribution.
What is the median distance between major cracks?
Let X be the distance between major cracks. Since =1/5=0.2, the pdf is f(x)=0.2e-0.2x , x>0

P(X>median)=0.5 for continuous data. P(X>median)=
 f ( x)dx  e
0.2 median
then
median
e 0.2 median  0.5 where median=-ln(0.5)/0.2=3.4657
3.
What is the probability that the distance between major cracks exceeds the mean value by more than 2
standard deviations?

P(X>+2)=P(X>5+2(5))=P(X>15)=
 0.2e
0.2 x
dx  e 0.2(15) =0.0498
15
4.
If you were counting the number of cracks on a circular region with radius 0.5 miles, which of the
following distributions you would use to find the probability of at least 2 cracks on that circular
region?
(a) Binomial
(b) Poisson
notice that the data are discrete
(c) Hypergeometric
(d) Exponential
5.
In which of the following cases, you can approximate the binomial distribution with the normal
distribution?
(a) n=100, p=0.03
(b) n=50, p=0.22
np10 and n(1-p)10
(c) n=10000, p=0.0009
6.
In which of the following cases, you can approximate the binomial distribution with the poisson
distribution?
(a) n=100, p=0.03
(b) n=50, p=0.22
(c) n=10000, p=0.0009
n and p0
7.
Calculate (9) .
(9-1)!=40320
8.
Calculate  (9 / 2)
(7 / 2)(5 / 2)(3 / 2)(1 / 2)(1 / 2)  105  / 16 =11.6325