Let [tex]\mu[/tex] be the mean of [tex]X[/tex]. Then
[tex]\mathbb P(X\le10)=0.1841\iff\mathbb P\left(\underbrace{\dfrac{X-\mu}{20}}_Z\le\dfrac{10-\mu}{20}\right)=0.1841[/tex]
For a random variable [tex]Z[/tex] following the standard normal distribution, we have
[tex]\mathbb P(Z\le z)=0.1841\implies z\approx-0.8999[/tex]
Transform the random variable to get this critical value in terms of [tex]\mu[/tex]:
[tex]\dfrac{10-\mu}{20}=-0.8999\implies\mu\approx27.997[/tex]
