The health center of a small, private university requires students to have appointments for visits. Each appointment takes an average of minutes with a standard deviation of minutes. The health center makes appointments each day and is scheduled to be open for hours. What is the probability that the total time spent for students who have appointments for tomorrow will exceed hours? (Hint: hours minutes.) Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Some data in the question are missing. The complete question is :

The health center of a small, private university requires students to have appointments for visits. Each appointment takes an average of 5.6 minutes with a standard deviation of 3.5 minutes. The health center makes 100 appointments each day and is scheduled to be open for 10 hours. What is the probability that the total time spent for 100 students who have appointments for tomorrow will exceed 10 hours? (Hint: 10 hours=600 minutes.) Carry your intermediate computations to at least four decimal places. Report your result to at least three decimal places.

Solution :

Let the time spent by [tex]$i^{th} $[/tex] students for the appointment be = [tex]$X_i$[/tex]

∴ E([tex]$X_i$[/tex]) = 5.6

   V([tex]$X_i$[/tex]) = [tex]$3.5^2$[/tex]

n = 100

The required probability

[tex]$P\left( \sum_{i=1} ^{100} X_i \leq 600 \right)$[/tex]

Now, [tex]$E\left( \sum_{i=1} ^{100} X_i \right) = 100 \times E (X_i) $[/tex]

                            = 100 x 5.6

                            = 560

[tex]$V\left( \sum_{i=1} ^{100} X_i \right) = (100)^2 \times (3.5)^2 $[/tex]

Standard deviation = 3.5 x 100

                                = 35

[tex]$P\left( \sum_{i=1} ^{100} X_i \leq 600 \right)$[/tex]

[tex]$\Rightarrow P\left( \frac{\sum_{i=1}^{100} X_i -560} {35} \leq \frac{600-560}{35}\right)$[/tex]

[tex]$= \phi (1.14286)$[/tex]

= 0.8735

= 0.874  (round to 3 decimal places)

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