Question

The cycle times for a truck hauling concrete to a highway construction site are uniformly distributed over the interval 50 to 70 minutes.

Required:
What is the probability that the cycle time exceeds 65 minutes if it is known that the cycle time exceeds 55 minutes?

Answer 1

The probability that the cycle time exceeds 65 minutes if it is known that the cycle time exceeds 55 minutes, should be 1 / 3.

Explanation:

It is known that the cycle times for a truck hauling concrete is uniformly distributed over a time interval of ( 50, 70 ). If c = cycle time, according to the question the probability that the cycle exceeds 65 minutes, respectively exceed 55 minutes should be the following - '
`Probability( c > 65 | c > 55 )`. '

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`f( c ) = \left \{ {{1 / 20,} \atop {0}} \right. \\50< c<70 - ( elsewhere )`

We know that the formula for Probability( A | B ) is P( A ∩ B ) / P( B ),

`P( c > 65 | c > 55 ) =`
`P( c > 55` ∩
`c > 65 )` /
`P( c > 55 )`,

And now we come to the formula
`P( a < c < b )` =
`\int\limits^(70)_(65) {f(x)} \, dc`. Substitute known values to derive two solutions, forming a fraction that represents the probability we desire.

`P( 65<c<70) = \int\limits^(70)_(65) {f(y)} \, dy\\ = \int\limits^(70)_(65) {(1/20)} \, dy\\ \\= 0.25`

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`P( 55<c<70) = \int\limits^(70)_(55) {f(y)} \, dy\\ = \int\limits^(70)_(65) {(1/20)} \, dy\\ \\= 0.75`

Take 0.25 over 0.75, 0.25 / 0.75, simplified to the fraction 1 / 3, which is our solution.

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Probability: 1 / 3