The graph of this distribution is:
Now, from it we know that the probability has to be less 0.1586 but more than 0.0227.
To find the probability we have to use the z-score to transform our distribution to a standard normal distribution.
The z-score is given by:
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/h06hsre30elxbqnbdkqzw5pbp57988qa0r.png)
Then in this case we have:
![\begin{gathered} z=(131.9-101)/(20.4) \\ z=1.5147 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/pxhuopnkce7knoyx5rngjowzm44103uggu.png)
Then our probability is:
![P(X>131.9)=P(Z>1.5147)](https://img.qammunity.org/2023/formulas/mathematics/college/1hb8xglubbjg228qsx0vsci9v9p9409oug.png)
Looking on a table for the value of the probability we have that:
![P(X>131.9)=P(Z>1.5147)=0.0649](https://img.qammunity.org/2023/formulas/mathematics/college/msuw0a3ylj36h3z4614b71s7kolyj25fvx.png)
Therefore the probability is 0.0649; which is the same as 6.49%.