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.
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?
4 answers
A) p(-2<z<-0.67）=0.2287
Please do the next or b
Do the same to b