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
Tossing two coins twice. I’m guessing the H and T stands for heads and tails.
8 centimeters equals to 80 millimeters
THE ANSWER YOU LOOKING FOR IS 21G
Answer:
12 cups oats, 4 cups raisins, 8 cups nuts
Step-by-step explanation:
For this problem, just multiply each ingredient by 4
