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?.
a) Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
b) Z = (score-mean)/SEm
SEm = SD/√n
Use same table.
To find the probability in both scenarios, we can use the concept of the standard normal distribution. In order to do this, we need to standardize the values using the z-score formula:
a) For a single bottle:
First, calculate the z-score using the formula:
z = (x - μ) / σ
where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
Substituting the values:
z = (296 - 300) / 3 = -4 / 3 = -1.33
Next, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator.
b) For the average of six bottles:
Since we are calculating the average, we need to take into account the standard error of the mean (SE) which is given by σ / √n, where n is the sample size.
For this scenario, n = 6, σ = 3, and x remains 296.
Calculate the standard error:
SE = 3 / √6 ≈ 1.22
Next, we calculate the z-score using the formula:
z = (x - μ) / SE
Substituting the values:
z = (296 - 300) / 1.22 ≈ -3.28
Again, we need to find the probability corresponding to this z-score using a standard normal distribution table or a calculator.
Note: The standard normal distribution table provides the probabilities for z-scores. If you don't have access to a table, you can use a calculator or an online calculator that can compute these probabilities for you.