Nicholas1998
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A large random sample of soda cans found that the volume of soda in a can had a normal distribution with a mean volume of 12 fl oz of soda and a standard deviation of 0.11 fl oz. Based on this, what percent of soda cans would have a volume less than 11.8 fl oz?

Use the standard normal table to look up the percent. Put your answer in percent form, and round to two decimal places. Do not include the percent symbol.

Sagot :

raphealnwobi

3.51 percent of soda cans would have a volume less than 11.8 fl oz.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\Where\ x\ is\ raw \ score,\mu=mean,\sigma=standard\ deviation\\\\\\Given\ that\ \mu=12,\sigma=0.11:For\ x<11.8\\\\z=\frac{11.8-12}{0.11}=-1.81\\\\[/tex]

From the normal distribution table, P(x < 11.8 ) = P(z < -1.81) = 0.0351 = 3.51%

Therefore, 3.51 percent of soda cans would have a volume less than 11.8 fl oz.

Find out more at: https://brainly.com/question/7449180.

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