Smallest value for n for a probability

The random variable y has a N(40,12) distribution. The mean of a random sample of n observations of y is denoted by X.
Find the smallest value of n for which P(x>=41) <0.05
How do you do this? Using proportions? :S

X is N(40,12/n)
P[X >= 41] = P[(X-40)/sqrt(12/n) >= (41-40)/sqrt(12/n)] < 0.05
P[Z >= (41-40)/sqrt(12/n)] < 0.05
(41-40)/sqrt(12/n) >= 1.645 (Table value)
sqrt(12/n) < 1/1.645
12/n < 0.3695
n >12/0.3695
n > 32.48
Sample size should be 33 or more.

the z-score for a right tail probability of 0.05 is 1.645
standard error = sqrt(12/n), so
(41-40)/sqrt(12/n) > 1.645, which yields
n > 12*1.645^2 >32.47
=> 33
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