What is this question asking (Normal distribution)

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
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If is asking to find P(X>2Y)? if so, how do i do that, do i replace with X with X=X-2Y, into Z

Let W = X-2Y
Mean of W = mean of X - 2 Mean of Y = -4
Variance = standard deviation squared
Variance of W = V(X-2Y) = V(X)+4V(Y) = 2.25+4(1.44) =8.01
Standard deviation of Z = sqrt(8.01) = 2.8302

P( X > 2Y) = P( X-2Y > 0) = P(W > 0)

μ = -4
σ = 2.8302
standardize w to z = (x - μ) / σ
P(w > 0) = P( z > (0--4) / 2.8302)
= P(z > 1.4133) =0.0793
(From Normal probability table)