We know that the probability in a normal distribution can be obtained by the z score, the z score is given as:
[tex]Z=\frac{x-\mu}{\sigma}[/tex]where mu is the mean and sigma is the standard deviation. In this case we have:
[tex]\begin{gathered} P(X<2.9)=0.2 \\ P(Z<\frac{2.9-3.34}{\sigma})=0.2 \\ P(Z<\frac{-0.44}{\sigma})=0.2 \end{gathered}[/tex]Now, to have 20% (that is 0.2) we need that the z score to be -0.842, then we have:
[tex]\begin{gathered} -\frac{0.44}{\sigma}=-0.842 \\ \sigma=\frac{-0.44}{-0.842} \\ \sigma=0.52256532 \end{gathered}[/tex]Therefore the standard deviation is $0.52
