<p>A doctor is on duty at a certain clinic. The time taken to attend to a patient is assumed to follow a normal distribution with mean $15$ minutes and standard deviation $3$ minutes. There is no wait time between patients. Suppose at $8$am, three patients arrive at the clinic and the first patient is immediately attended to.</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the probability that the second patient has to wait longer than $10$ minutes.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Let $M$ denote the shortest time, in minutes, attended by the doctor among the first three patients. Find $\mathrm{P}\left( M>10 \right)$.</span></div><p style="text-align: right;">[1]</p></li><li><div class="list-item-content"><span class="list-item-text">Let $L$ denote the longest time, in minutes, attended by the doctor among the first three patients. Find $\mathrm{P}\left( L<10 \right)$.</span></div><p style="text-align: right;">[1]</p></li></ol></li><li><div class="list-item-content"><span class="list-item-text"></span></div><ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the probability that the third patient is attended by the doctor before $8.30$am on the same day.</span></div><p style="text-align: right;">[2]</p></li><li><div class="list-item-content"><span class="list-item-text">Given that the doctor took a total of at least $25$ minutes to attend to the first two patients, find the probability that the third patient is attended by the doctor before $8.30$am on the same day.</span></div><p style="text-align: right;">[3]</p></li></ol></li><li><div class="list-item-content"><span class="list-item-text">After lunch, an additional doctor is on duty and she is able to attend to patients twice as quickly as the first doctor.Assume that patients will be attended by the earliest available doctor in the clinic and that there is no wait time between patients.If there are three patients waiting to be attended by either doctor, find the probability that the third patient is attended by the first doctor.</span></div><p style="text-align: right;">[3]</p></li></ol>

A doctor is on duty at a certain clinic. The time taken to attend to a patient is assumed to follow a normal distribution with mean $15$ minutes and standard deviation $3$ minutes. There is no wait time between patients. Suppose at $8$am, three patients arrive at the clinic and the first patient is immediately attended to.

Find the probability that the second patient has to wait longer than $10$ minutes.

Let $M$ denote the shortest time, in minutes, attended by the doctor among the first three patients. Find $\mathrm{P}\left( M>10 \right)$.

Let $L$ denote the longest time, in minutes, attended by the doctor among the first three patients. Find $\mathrm{P}\left( L<10 \right)$.

Find the probability that the third patient is attended by the doctor before $8.30$am on the same day.

Given that the doctor took a total of at least $25$ minutes to attend to the first two patients, find the probability that the third patient is attended by the doctor before $8.30$am on the same day.

After lunch, an additional doctor is on duty and she is able to attend to patients twice as quickly as the first doctor.Assume that patients will be attended by the earliest available doctor in the clinic and that there is no wait time between patients.If there are three patients waiting to be attended by either doctor, find the probability that the third patient is attended by the first doctor.

2012 HCI C2 BT2 P2 Q4

2012 HCI C2 BT2 P2 Q4

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