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A physiotherapist has a patient doing some slow jogging on a treadmill and sets the timer for 24 minutes. A little while later, the physiotherapist’s assistant comes over to check on the patient, not knowing how much time is left on the treadmill.

a. If the speed needs to be adjusted with 11 minutes left to go, what is the probability that the assistant will need to adjust the speed of the treadmill? P(X<11)

b. How much time would the assistant expect to be remaining on the timer? E(X)= µ =

User Michael J
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Final answer:

The probability that the assistant needs to adjust the speed of the treadmill with 11 minutes left is P(X<11) = 11/24. The expected time remaining on the treadmill (E(X) = µ) would be 12 minutes, assuming any time during operation is equally likely for a check to occur.

Step-by-step explanation:

The student is asking about probability and expected value, two fundamental concepts in statistics. Since the problem setup is related to a continuous event (the time remaining on the treadmill), we can consider probability in terms of a uniform distribution, as we're given no further information on the timing of the adjustments. To address part a), if the total time set is 24 minutes, and the assistant checks at a random time, the probability that there are fewer than 11 minutes left (P(X<11)) would essentially be the time left (11 minutes) divided by the total time (24 minutes), assuming any moment during the treadmill's operation is equally likely for the check to happen.

For part b), the expected time remaining on the treadmill's timer, without knowing the exact distribution of when the physiotherapist's assistant checks the timer, we would take the average time remaining, which in this case is simply the middle point of the 24-minute period. So, the expected time remaining would be half of 24 minutes, which is 12 minutes.

User Clint B
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