The Solution:
The formula for Z score is
![Z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}](https://img.qammunity.org/2023/formulas/mathematics/college/ydwyzc6uwe40cm2qrw358g78bvf6zmltbq.png)
In this case,

Substituting these values in the formula, we get the Z-score.
![Z=\frac{\bar{0.815}-0.904}{\frac{0.299}{\sqrt[]{50}}}=\frac{\bar{0.815}-0.904}{\frac{0.299}{\sqrt[]{50}}}=(-0.089)/(0.0422849855)=-2.104766](https://img.qammunity.org/2023/formulas/mathematics/college/gr11kcxg6n58u3ks2oard0rhoe1d4jfny9.png)
From the Z-distribution tables, we have

Thus, the probability of randomly selecting 50 cigarettes with a mean of 0.815g or less is 0.017656.