Answer:
6.68% or 0.0668
Step-by-step explanation:
Mean sheet length (μ) = 30.05 inches
Standard deviation (σ) = 0.2 inches
In a normal distribution, for any length X, the z-score is determined by the following expression:
[tex]z=\frac{X-\mu}{\sigma}[/tex]
For X = 29.75, the z-score is:
[tex]z=\frac{29.75-30.05}{0.2}\\z=-1.5[/tex]
A z-score of -1.5 corresponds to the 6.68th percentile of a normal distribution.
Therefore, the probability that a sheet selected at random will be less than 29.75 inches long is 6.68%.
