A manufacturer of cream filled donuts wants to make sure that it's automatic filling process is on target. Based On sample of 25 donuts the mean weight of cream filling is estimated to be 15 grams with a standard deviation of 1.5 grams. however, the quality control inspectors wish to estimate the mean wright of cream filling mor accurately so that they can be 99% confident that it is within 0.25 grams of the true mean. How many donuts should they sample?

1

step-by-step explanation:

you are throwing one dart

answer: your answer is a

irst, find the z score of a score of 80. this is given by the formula z = (x-u)/o, where x is the score, u is the mean, and o is the standard deviation. this gives

z = (80-78)/12 = 1/6 = .1667

now, find what percentile a z score of 1/6 correlates to in the normal distribution.

a z score of 1/6 = .1667 is equivalent to the 56.36th percentile of the distribution. so the probability a score is over this is 1-.5636 = .4364

the mean of a sample mean is equal to the original mean. however, the standard deviation of a sample mean is equal to the original standard deviation divided by the square root of n (sample size). so the new standard deviation is 12/sqrt(16) = 3.

the new z score is 80-78/3 = 2/3 = .667. this is equivalent to the 74.86th percentile. so the probability that the mean will be greater than this score is 1-.7486 = .2514.

yes

step-by-step explanation:

lol