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# In a company dedicated to packaging beer in 750 mL containers, a normal distribution is handled in its packaging process, which registers an average of 745 mL and a standard deviation of 8 mL. Determine: a) The probability that a randomly selected container exceeds 765 mL of beer b) The probability that the beer content of a randomly selected container is between 735 and 755 mL.

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## Answer to a math question In a company dedicated to packaging beer in 750 mL containers, a normal distribution is handled in its packaging process, which registers an average of 745 mL and a standard deviation of 8 mL. Determine: a) The probability that a randomly selected container exceeds 765 mL of beer b) The probability that the beer content of a randomly selected container is between 735 and 755 mL.

Darrell
4.5
63 Answers
To solve this problem, we will use the properties of the normal distribution.

a) To find the probability that a randomly selected container exceeds 765 mL of beer, we need to calculate the z-score corresponding to 765 mL using the formula:

z = \frac{{X - \mu}}{{\sigma}}

where X is the value we are interested in $765 mL$, μ is the mean $745 mL$, and σ is the standard deviation $8 mL$. Substituting these values, we get:

z = \frac{{765 - 745}}{{8}} = \frac{{20}}{{8}} = 2.5

Next, we need to find the area under the normal curve to the right of the z-score of 2.5. We can use a standard normal distribution table or a calculator to find this area.

Using a standard normal distribution table, we find that the area to the right of 2.5 is approximately 0.0062.

Therefore, the probability that a randomly selected container exceeds 765 mL of beer is approximately 0.0062.

b) To find the probability that the beer content of a randomly selected container is between 735 and 755 mL, we need to calculate the z-scores corresponding to 735 mL and 755 mL.

For 735 mL:
z_1 = \frac{{735 - 745}}{{8}} = \frac{{-10}}{{8}} = -1.25

For 755 mL:
z_2 = \frac{{755 - 745}}{{8}} = \frac{{10}}{{8}} = 1.25

Next, we need to find the area under the normal curve between these two z-scores. We can subtract the area to the left of -1.25 from the area to the left of 1.25. Again, we can use a standard normal distribution table or a calculator to find these areas.

Using a standard normal distribution table, we find that the area to the left of -1.25 is approximately 0.1056, and the area to the left of 1.25 is approximately 0.8944.

Therefore, the probability that the beer content of a randomly selected container is between 735 and 755 mL is approximately 0.8944 - 0.1056 = 0.7888.

Answer:
a) The probability that a randomly selected container exceeds 765 mL of beer is approximately 0.0062.
b) The probability that the beer content of a randomly selected container is between 735 and 755 mL is approximately 0.7888.

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