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S3 help! urgent

Cola cans are delivered to shops in boxes of 24 cans, and are normally distibuted with mean weight 350g with a variance of 8g. the weights of empty boxes are normally distributed with mean 100g and standard deviation 2g.

Find the probability that a full box of cola cans weighs between 8.51kg and 8.52kg.

answer in the back says 0.1625 but I have no clue how this result was got.

Really would appreciate some help!!
Original post by fg(x)
Cola cans are delivered to shops in boxes of 24 cans, and are normally distibuted with mean weight 350g with a variance of 8g. the weights of empty boxes are normally distributed with mean 100g and standard deviation 2g.

Find the probability that a full box of cola cans weighs between 8.51kg and 8.52kg.

answer in the back says 0.1625 but I have no clue how this result was got.

Really would appreciate some help!!


Let XX denote the r.v. representing the can weight, and YY denote the r.v. representing the box weight.

We have;

X∼N(350,8)X \sim N(350,8)
Y∼N(100,2)Y \sim N(100,2)

The random variable ZZ representing a full product weight would include 24 cans and a box. We can say that Z=24X+YZ=24X+Y

You are looking for P(8510<Z<8520)P(8510 < Z < 8520)

Have a go now.
Original post by fg(x)
Cola cans are delivered to shops in boxes of 24 cans, and are normally distibuted with mean weight 350g with a variance of 8g. the weights of empty boxes are normally distributed with mean 100g and standard deviation 2g.

Find the probability that a full box of cola cans weighs between 8.51kg and 8.52kg.

answer in the back says 0.1625 but I have no clue how this result was got.

Really would appreciate some help!!


As RDKgames said, except you want Z=∑i=124Xi+Y\displaystyle Z=\sum_\text{i=1}^{24}X_i+Y, rather than 24X+Y, since the cans are 24 instances rather than 24 times one instance.

This does not effect the mean, but it does effect the variance of Z.
Original post by ghostwalker
Z=∑i=124Xi+Y\displaystyle Z=\sum_\text{i=1}^{24}X_i+Y, rather than 24X+Y


Ah, should've seen my error there - doesn't help that I have an exam on this stuff soon :lol:
Reply 4
Avatar for fg(x)
fg(x)
OP
7
Original post by ghostwalker
As RDKgames said, except you want Z=∑i=124Xi+Y\displaystyle Z=\sum_\text{i=1}^{24}X_i+Y, rather than 24X+Y, since the cans are 24 instances rather than 24 times one instance.

This does not effect the mean, but it does effect the variance of Z.


Yes this is what confused me. I'm right in doing 24 lots of the variance of X not 24 squared?
Reply 5
Avatar for thefreys
thefreys
5
Original post by RDKGames
Ah, should've seen my error there - doesn't help that I have an exam on this stuff soon :lol:


wow you're only doing a levels? nice
Original post by thefreys
wow you're only doing a levels? nice


No. I have a uni module which has this stuff as part of its spec
Original post by fg(x)
Yes this is what confused me. I'm right in doing 24 lots of the variance of X not 24 squared?


That's correct. Just 24 lots.

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