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Sagot :
Given data:
Mean: 993mL
Standard deviation: 7mL
Find p(988
1. Find the z-value corresponding to (x>988), use the next formula:
[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ \\ z=\frac{988-993}{7}=-0.71 \end{gathered}[/tex]2. Find the z-value corresponding to (x<991):
[tex]z=\frac{991-993}{7}=-0.29[/tex]3. Use a z score table to find the corresponding values for the z-scores above:
For z=-0.71: 0.2389
For x=-0.29: 0.3859
4. Subtract the lower limit value (0.2389) from the upper limit value (0.3859):
[tex]0.3859-0.2389=0.147[/tex]5. Multiply by 100 to get the percentage:
[tex]0.147*100=14.7[/tex]Then, 14.7% of the bottles have volumes between 988mL and 991mL

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