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A bottler of drinking water fills plastic bottles with a mean volume of 993 milliliters (mL) and standard deviation of 7 mL. The fill volumes are normally distributed. What proportion of bottles have volumes between 988 mL and 991 mL?

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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