need to know to do this question is that if X and Y are random vairables with
means mx, and my, and variances vx and vy, then the mean and variance of
Z=aX+bY are a a.mx+b.my and a^2.vx+b^2.vy, and if X and Y are normally
distributed then so is Z, where a and b are real numbers.
In this case X and Y are the weights of the first and second small packets
chosen and a and b are both 1, and so their means and variances are equal
to 500g and 10^2 g^2 respectivly (the variance is the square of the standard deviation).
So the mean and variance of Z=X+Y are 1000g and 200 g^2 respectivly, and
so the mean and standard deviations are 1000g and sqrt(200) g.
The z-scores coresponfing to 990 g and 1020 g are: -10/sqrt(200), and
20/sqrt(200). So if P(x) is the cumulative distribution function of the
standard normal distribution:
p(990<x+y<1020) = P(20/sqrt(200)) - P(-10/sqrt(200)) ~= 0.921 - 0.261 = 0.66.
Here we want to mean and variance of the RV Z=A - B - C, where A is theb. find the probability that the weight of one randomly chosen large packet exceeds the total weight of two randomly chosen small packets by at least 25 gram.
weight of the large packet, B the weight of the first small packet, and C the
weight of the second small packet.
Then the mean and variance of Z are 0g, 15^2+10^2+10^2 g^2,so
the mean and sd are 0g and sqrt(425), and you are asked for p(z>25).
Here we put Z=0.5 A - B, where A is the weight of the large packet, and Bc. find the probability that one half of the weight of one randomly chosen large packet exceeds the weight of one randomly chosen small packet by at least 12.5 gram.
the weight of the small packet. Then the mean and sd of Z are 0,
sqrt(156.25), and the question asks for: