# Two very similar probability questions, just need a confirmation

1. Nov 13, 2013

### GandhiReborn

1. The problem statement, all variables and given/known data

Question 1)

A second hand bookshop owner notes that the weights (in gram) of paperbacks are Normally distributed with mean m1 and standard deviation s1, while the weights (again, in grams) of hardbacks are Normally distributed with mean m2 and standard deviation s2, where

m1 = 246, s1 = 16,
m2 = 306, s2 = 20,

Assume that the books are chosen at random.

Let x = 834.

Then the probability that one paperback and two hardback together weigh less than x grammes is:

Question 2)

A second hand bookshop owner notes that the weights (in grams) of paperbacks are Normally distributed with mean m1 and standard deviation s1, while the weights (again, in grams) of hardbacks are Normally distributed with mean m2 and standard deviation s2, where

m1 = 243, s1 = 12,
m2 = 303, s2 = 16,

Assume that the books are chosen at random.

Let M = 14, a = 4243, b = 4843.

Then the probability that a selection of M different hardbacks weighs between a and b grams is

2. The attempt at a solution

Question 1)

I said we would have to form a combined normal distribution, but first we must take the mean and variance of the paperbacks and multiply them by the number of books we are considering, 2.
Then we would add these values to the variance and mean of the hardbacks. This gives a new standard deviation = 32.496 (having sqrt'd the variance) and a new mean of 858.

Following this and using Excel, calculation would be simply "=normdist(834,858,32.496,true) to give us a probability of 0.23009 - is this correct?

Question 2)

I didn't understand whether both pieces of data was needed, so I just took the second set. Similar to before, I worked out the variance of that dataset, multiplied it by the number of books being considered - 14 - then took the new standard deviation. I also multiplied the mean by 14.

Then I just used the function =NORMDIST(4843,4242,59.867,TRUE)-NORMDIST(4243,4242,59.867,TRUE) and that gave me the final answer of 0.493.

Again, is this correct?

Thank you!

2. Nov 13, 2013

### Ray Vickson

3. Nov 13, 2013

### GandhiReborn

Awesome, massively appreciate the kind words and confirmation - thank you :D

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