We know that the probability in a normal distribution can be obtained by the z score, the z score is given as:
![Z=(x-\mu)/(\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/sv72d3baryltp7s92ka7mqqorx2970ht60.png)
where mu is the mean and sigma is the standard deviation. In this case we have:
![\begin{gathered} P(X<2.9)=0.2 \\ P(Z<(2.9-3.34)/(\sigma))=0.2 \\ P(Z<(-0.44)/(\sigma))=0.2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/1swr8428hltdt31wh79chs4y2znueu0s9e.png)
Now, to have 20% (that is 0.2) we need that the z score to be -0.842, then we have:
![\begin{gathered} -(0.44)/(\sigma)=-0.842 \\ \sigma=(-0.44)/(-0.842) \\ \sigma=0.52256532 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/w48upx20r4a8tha3i866tkyjb16h9xwvdz.png)
Therefore the standard deviation is $0.52