Soft-drink cans are filled by an automated filling machine and the standard deviation is 0.5 fluid ounces assume that the fill v
olumes of the cans are independent, normal random variables. (a what is the standard deviation of the average fill volume of 100 cans? (b if the mean fill volume is 12.1 ounces, what is the probability that the average fill volume of the 100 cans is below 12 fluid ounces? (c what should the mean fill volume equal so that the probability that the average of 100 cans is below 12 fluid ounces is 0.005? (d if the mean fill volume is 12.1 fluid ounces, what should the standard deviation of fill volume equal so that the probability that the average of 100 cans is below 12 fluid ounces is 0.005? (e determine the number of cans that need to be measured such that the probability that the average fill volume is less than 12 fluid ounces is 0.01.
A.) For n independent variates with the same
distribution, the standard deviation of their mean is the standard
deviation of an individual divided by the square root of the sample
size: i.e. s.d. (mean) = s.d. / sqrt(n) Therefore, the standard deviation of of the average fill volume of 100 cans is given by 0.5 / sqrt(100) = 0.5 / 10 = 0.05
b.) In a normal distribution, P(X < x) is given by P(z < (x - mean) / s.d). Thus, P(X < 12) = P(z < (12 - 12.1) / 0.05) = P(z < -2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275
c.) Let the required mean fill volume be u, then P(X < 12) = P(z < (12 - u) / 0.05) = 1 - P(z < (u - 12) / 0.05) = 0.005 P(z < (u - 12) / 0.05) = 1 - 0.005 = 0.995 = P(z < 2.575) (u - 12) / 0.05 = 2.575 u - 12 = 2.575 x 0.05 = 0.12875 u = 12 + 0.12875 = 12.12875 Therefore, the mean fill volume should be 12.12875 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.
d.) Let the required standard deviation of fill volume be s, then P(X < 12) = P(z <
(12 - 12.1) / s) = 1 - P(z < 0.1 / s) = 0.005 P(z < 0.1 / s) = 1 - 0.005 = 0.995 = P(z < 2.575) 0.1 / s = 2.575 s = 0.1 / 2.575 = 0.0388 Therefore, the standard deviation of fill volume should be 0.0388 so that the probability that the average of 100 cans is below 12 fluid ounces be 0.005.
e.) Let the required number of cans be n, then P(X < 12) = P(z <
(12 - 12.1) / (0.5/sqrt(n))) = 1 - P(z < (12.1 - 12) / (0.5/sqrt(n))) = 0.01 P(z < 0.1 / (0.5/sqrt(n))) = 1 - 0.01 = 0.99 = P(z < 2.327) 0.1 / (0.5/sqrt(n)) = 2.327 0.5/sqrt(n) = 0.1 / 2.327 = 0.0430 sqrt(n) = 0.5/0.0430 = 11.635 n = 11.635^2 = 135.37 Therefore, the number of cans that need to be measured such that the average fill volume is less than 12 fluid ounces be 0.01
