Question

A beverage company produces cans of lemonade. According to the label a can contains a volume of 375mL of lemonade. A consumer group tested this claim by measuring the volume of lemonade in 2,500 cans. They found that the volume was normally distributed with a standard deviation of 5mL and a mean of 372mL.
If a can is chosen at random, find:

P (its volume is ≥382mL)

Answer 1

The probability that the volume of a can chosen at random is ≥382 is 0.47725

Normal distribution probability

Probability is the likelihood or chance that an event will occur. For the given question, we need to calculate the  z value as shown;

$z=\dfrac{x-\mu}{\sigma}$

Given the following parameters

$x=382mL\\\mu (mean)=372mL\\\sigma(S.D)=5mL$

Substitute

$z=\dfrac{382-372}{5} \\z=\dfrac{10}{5}\\ z=2$

In order to calculate the probability that the volume of a can chosen at random is ≥382, we will look for (z≥2)

z≥2 according to probability table is equivalent to 0.47725

Learn more on normal distribution here: brainly.com/question/26822684

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