Find z such that 36.9% of the standard normal curve lies to the right of z. Shade the curve (by hand).
P(Z>z0) = .369
So, by NORMSINV(1 - 0.369) forumula in excel we have :
| z0 = +0.3345 |
So, the shaded area to the right of z0 (marked in blue below) = + 0.3345 which has a probability of 36.9% is:

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