In a chemical process, the amount of a certain type of impurity in the output is difficult to control and is thus a randomvariable. speculation is that the population mean amount of the impurity is 0.20 gram per gram of output. it is known thatthe standard deviation is 0.1 gram per gram. an experiment is conducted to gain more insight regarding the speculationthat u = 0.2 the process is run on a lab scale 50 times and the sample average 3% turns out to be 0.23 gram per gram. comment on the speculation that the mean amount of impurity is 020 gram per gram. make use of the central limittheorem in your work.

well the answer to that is xoxoxo

step-by-step explanation:

there is no graph so i cant sorry

step-by-step explanation:

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.

this cant be answered without the graph. you may need to add it.