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In: Statistics and Probability

The distance between major cracks in a highway follows an exponential distribution with a mean of...

The distance between major cracks in a highway follows an exponential distribution with a mean of 21 miles. Given that there are no cracks in the first five miles inspected, what is the probability that there is no major crack in the next 5 mile stretch? Please enter the answer to 3 decimal places.

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Expert Solution

SOLUTION:

From given data,

From the information, observe that the distance between major crackers in a highway follows an exponential distribution with mean of 21 miles .

Consider X is the random variable that represents the distance between major cracks in a highway.

The random variable X follows an exponential distribution

The density function is as follows

= >0

The distribution function is as follows:

F(x) = 1 -

The mean number of miles is, =21.

Calculate parameter

= 1 / = 1/21 = 0.047

Given that there are no cracks in the first five miles inspected, what is the probability that there is no major crack in the next 5 mile stretch?

Calculate the probability that there are no major cracks in the next 5 miles inspected given that there are no cracks in the first five miles inspected.

P(X >10 | X>5) = P(X > 5+5 | X>5)

= P(X > 5)

= 1 - P(X < 5)

= 1 - [1-]

= 0.79057

0.791 (answer to 3 decimal places)

Therefore, the probability that there are no major cracks in the next 5 miles inspected given that there are no cracks in the first five miles stretch is 0.791

orchestra answered 1 year ago

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