# Math Help - Linear combinations of random variable

1. ## Linear combinations of random variable

Small packets of nails are advertised as having average weight 500 g, and large packets as having average weight 1000 g. Assume that the packet weight are distributed normally with means as advertised, and standard deviations of 10 g for a small packet and 15 gram for a large packet. Giving your answers correct to 3 decimal places.

a. find the probability of two randomnly chosen packets have a total weight between 990 gram and 1020 gram.

b. 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.

c. 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.

2. Originally Posted by kingkaisai2

Small packets of nails are advertised as having average weight 500 g, and large packets as having average weight 1000 g. Assume that the packet weight are distributed normally with means as advertised, and standard deviations of 10 g for a small packet and 15 gram for a large packet. Giving your answers correct to 3 decimal places.

a. find the probability of two randomnly chosen packets have a total weight between 990 gram and 1020 gram.
Lets assume you mean two randomly chosen small packets. What you
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.

b. 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.
Here we want to mean and variance of the RV Z=A - B - C, where A is the
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).

c. 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.
Here we put Z=0.5 A - B, where A is the weight of the large packet, and B
the weight of the small packet. Then the mean and sd of Z are 0,
sqrt(156.25), and the question asks for:

p(z>=12.5).

RonL

3. ## Re: Linear combinations of random variable

I was just answering this same question RIGHT NOW.

I also need help regarding this.

Why is it that the variance is sqr. root of 200 and not sqr root of 400? THIS IS WHAT I DO NOT UNDERSTAND.

When do we use : a^2 x Var(X) AND when do we use a x Var(X) ???

Please explain. Would highly appreciate an answer today. Have my finals tomorrow !