Cola Corporation produces Orange Cola. The filling machines are adjusted to pour 12 ounces of soda into each 12-ounce can. However, the actual amount of soda poured into each can is not exactly 12 ounces; it varies from can to can. It has been observed that the net amount of soda in such a can has a normal distribution with a mean of 12 ounces and a standard deviation of 0.015 ounce.
a. What is the probability that a randomly selected can of Orange Cola contains between 11.97 to 11.99 ounces of soda?
b. What percentage of the Orange Cola cans contain 12.02 to 12.07 ounces of soda?
A) p(-2<z<-0.67）=0.2287
Please do the next or b
You have to find the z-score for each of the given values.
11.97 - 12 divided by 0.015 will give you a z. Do the same for 11.99.
You have to use a z-table or a calculator to find the the probability between those to z-scores.
Do the same for part b.
Do the same to b
To find the probabilities in these questions, we need to use the properties of the normal distribution. The normal distribution is defined by its mean (μ) and standard deviation (σ).
a. To find the probability that a randomly selected can of Orange Cola contains between 11.97 to 11.99 ounces of soda, we first need to standardize the values using the z-score formula.
The z-score formula is given by:
z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For 11.97 ounces of soda:
z1 = (11.97 - 12) / 0.015
Similarly, for 11.99 ounces of soda:
z2 = (11.99 - 12) / 0.015
Next, we need to find the probabilities corresponding to these z-scores using a z-table or a calculator that provides the area under the normal curve.
The probability of a can containing between 11.97 to 11.99 ounces of soda is the difference between the probabilities corresponding to z1 and z2:
P(11.97 ≤ X ≤ 11.99) = P(z1 ≤ Z ≤ z2)
b. To find the percentage of Orange Cola cans containing 12.02 to 12.07 ounces of soda, we follow a similar process.
Again, we need to standardize the values using the z-score formula and find the corresponding probabilities using a z-table or calculator.
For 12.02 ounces of soda:
z3 = (12.02 - 12) / 0.015
Similarly, for 12.07 ounces of soda:
z4 = (12.07 - 12) / 0.015
The percentage of cans containing between 12.02 to 12.07 ounces of soda is the difference in probabilities:
P(12.02 ≤ X ≤ 12.07) = P(z3 ≤ Z ≤ z4)
By finding the respective probabilities, we can determine the answers to both questions.