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Suppose the times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 13 minutes and 20 minutes. What is the probability that a randomly selected cable repair visit will take at least 19 minutes? (Enter the final answers reduced to the lowest possible fraction.)

User DirectX
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Answer:
\mathbf{(1)/(7)}

Explanation:

The probability density function for x which is uniformally distributed in interval [a,b] is given by :-


f(x)=(1)/(b-a)

Given : The times required for a cable company to fix cable problems in its customers' homes are uniformly distributed between 13 minutes and 20 minutes.

Then,
f(x)=(1)/(20-13)=(1)/(7)

Now, the probability that a randomly selected cable repair visit will take at least 19 minutes will be :_


\int^(20)_(19)\ f(x)\ dx\\\\=\int^(20)_(19)((1)/(7))\ dx\\\\= (1)/(7)[x]^(20)_(19)\\\\=(1)/(7)[20-19]=(1)/(7)

Hence, the probability that a randomly selected cable repair visit will take at least 19 minutes =
\mathbf{(1)/(7)}

User Zarsky
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