Bottles of a popular cola contain a mean amount of 300 mL of cola. There is some variation from bottle to bottle because the filling machinery is not perfectly precise. The distribution of the contents is normal with a standard deviation of 3 mL.

a) What is the probability that one randomly selected bottle will contain less than
296 mL of cola?

b) What is the probability that the average of 6 randomly selected bottles will contain less than 296 mL of cola?.

Z= x-u/o

So,
296-300/3
= -.77
Look at a standard normal distribution table
.2794 is the value from the table.
If you draw the curve, you will know that the area is to the left.
So, you subtract:
.5000- .2794
=.2206 is the probability.
I am not sure about the second part.

b) Z = (score-mean)/SEm

SEm = SD/√n

Use same table.

To answer these questions, we will use the concept of the normal distribution and z-scores.

a) To find the probability that one randomly selected bottle will contain less than 296 mL of cola, we need to calculate the z-score and use a z-table. The z-score is calculated as follows:

z = (X - μ) / σ

Where X is the value we are interested in (296 mL), μ is the population mean (300 mL), and σ is the standard deviation (3 mL).

Substituting the values into the formula, we get:

z = (296 - 300) / 3
z = -4 / 3
z = -1.33

Now we need to find the probability corresponding to this z-score in the standard normal distribution table (z-table). Looking up the z-score -1.33 in the z-table, we find that the corresponding probability is 0.0918.

Therefore, the probability that one randomly selected bottle will contain less than 296 mL of cola is approximately 0.0918 or 9.18%.

b) To find the probability that the average of 6 randomly selected bottles will contain less than 296 mL of cola, we need to calculate the z-score using the formula for the sample mean:

z = (X - μ) / (σ / √n)

Where X is the value we are interested in (296 mL), μ is the population mean (300 mL), σ is the standard deviation (3 mL), and n is the sample size (6 bottles).

Substituting the values into the formula, we get:

z = (296 - 300) / (3 / √6)
z = -4 / (3 / √6)
z = -4 / (3 / 2.449)
z = -4 / 1.225
z = -3.27

Looking up the z-score -3.27 in the z-table, we find that the corresponding probability is approximately 0.00056 or 0.056%.

Therefore, the probability that the average of 6 randomly selected bottles will contain less than 296 mL of cola is approximately 0.00056 or 0.056%.