Question

Sodas in a can are supposed to contain an average of 12 ounces. This particular brand has a standard deviation of 0.1 ounces, with an average of 12.1 ounces. If the can's contents follow a Normal distribution, what is the probability that the mean contents of a six pack are less than 12 ounces?

Answer 1

To find the probability that the mean contents of a six pack are less than 12 ounces, you need to calculate the z-score and use the z-table to find the corresponding probability. In this case, the probability is approximately 0.2912, or 29.12%.

Step-by-step explanation:

To find the probability that the mean contents of a six pack are less than 12 ounces, we need to calculate the z-score and then use the z-table to find the corresponding probability.

The formula to calculate the z-score is z = (x - μ) / (σ / sqrt(n)), where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

In this case, x = 12, μ = 12.1, σ = 0.1, and n = 6.

Plugging these values into the formula, we get z = (12 - 12.1) / (0.1 / sqrt(6)) = -0.5477.

Using the z-table, we can find that the corresponding probability for a z-score of -0.5477 is approximately 0.2912.

Therefore, the probability that the mean contents of a six pack are less than 12 ounces is 0.2912, or 29.12%.

Answer 2

The probability that the mean contents of a six pack are less than 12 ounces is 15.87%.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X.

In this problem, we have that:

`\mu = 12.1, \sigma = 0.1`

This probability is the pvalue of the zscore of
`X = 12`

`Z = (X - \mu)/(\sigma)`

`Z = (12 - 12.1)/(0.1)`

`Z = -1`

Z = 1 has a pvalue of .1587.

This means that the probability that the mean contents of a six pack are less than 12 ounces is 15.87%.