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.

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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
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.

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