How to solve this normal distribution question with 2 variables

The time X it takes a consultant to see a new patient at a hospital out-patient clinic may be assumed to be normally distributed with mean “16 minutes” and standard deviation “1.5 minutes”. The time Y it takes the same consultant to see a returning patient may be assumed to be normally distributed with mean “10 minutes” and standard deviation “1.2 minutes”. The consultation times of different patients mat be assumed to be independent.

(i)What is the probability the consultation time for the new patients is more than twice that of a returning patient

Answer: p=0.0793

(ii) At the start of the Monday clinic, 5 new patients and 8 returning patients are waiting to be seen by the consultant. What is the probability that the consultant will see all of them within 3 hours?
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(ii)i wanna know answer for ii, Do i have to use binomial distribution on 2. Can somone tell me how to solve it

Call the total time T

E[T] = 5E[X] + 8E[Y] = 5*16 + 8*10 = 160
Var[T] = Var[5X] + Var[8Y] = (5*1.5)^2 + (8*1.2)^2 = 148.41

To get the variance I used the fact that Cov[X,Y] = 0 since you said the times for all patients are independent.

So now you find the Z score.

Z = {T - E[T]}/StDev[T] = (T - 160)/sqrt(148.41)

So if T under 3 hours = 180 minutes is equivalent to

Z < 20/sqrt(148.41) = 1.6417

Looking up a standard normal table would give you a probability of 94.97%

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