High School Maths

The amount of jam in a jar has a normal distribution with mean 340g and standard deviation 10g.
The mass of the jar has a normal distribution with mean 150g and standard deviation 8g.

Find the probability that a randomly chosen pack of 20 jars of jam has a total mass exceeding 10000g

Solution 1
mean = 20*(340 + 150) = 9800g
SD =(20*10^2 + 20*8^2)^0.5 = 57.27g

P(X>10000) = P( Z>(10000 - 9800)/57.27) = P( Z>3.492)

Probability = 0.024%


Solution 2
mean = 20*(340 + 150) = 9800g
SD =((20*10)^2 + (20*8)^2)^0.5 = 256.125g

P(X>10000) = P( Z>(10000 - 9800)/256.125) = P( Z>0.781)

Probability = 21.75%



The problem here is to find the correct SD, should we use SD(x+x....+y+y) or SD(20x +20y) ? Please help.

I think solution 1 is correct.

Quote: ssho88

The amount of jam in a jar has a normal distribution with mean 340g and standard deviation 10g.
The mass of the jar has a normal distribution with mean 150g and standard deviation 8g.

Find the probability that a randomly chosen pack of 20 jars of jam has a total mass exceeding 10000g

Solution 1
mean = 20*(340 + 150) = 9800g
SD =(20*10^2 + 20*8^2)^0.5 = 57.27g

P(X>10000) = P( Z>(10000 - 9800)/57.27) = P( Z>3.492)

Probability = 0.024%


Solution 2
mean = 20*(340 + 150) = 9800g
SD =((20*10)^2 + (20*8)^2)^0.5 = 256.125g

P(X>10000) = P( Z>(10000 - 9800)/256.125) = P( Z>0.781)

Probability = 21.75%



The problem here is to find the correct SD, should we use SD(x+x....+y+y) or SD(20x +20y) ? Please help.

I think solution 1 is correct.

You're right--solution 1 is correct. The variance of a sum of uncorrelated random variables equals the sum of their variances.

So, the variance of the mass of a collection of 20 jams and 20 jars would be:

Var. = 10² + 10² + 10² + 10² +10² + 10² +10² + 10² +10² + 10² +10² + 10² +10² + 10² +10² + 10² +10² + 10² +10² + 10² + 8² + 8² + 8² + 8² +8² + 8² +8² + 8² +8² + 8² +8² + 8² +8² + 8² +8² + 8² +8² + 8² +8² + 8²
= 20*10² + 20*8² = 20*(10²+8²)
= 3280 g²

S.D. = 3280^0.5 = 57.27 g.

Solution 2 is wrong because it is just using 20 as a scaling factor, as if you were changing units. For example, if you wanted to change your units from grams to a new unit 1/20 as big, the variance for a jar would change from 8² to (20²)(8²).

That wasn't taught in my high school

Quote: darkoz

That wasn't taught in my high school

It wasn't taught in my high school either. I'm from the U.S., but the original poster is from another country as revealed by his calling math "maths." And I bet most countries have higher math education standards than the United States.

I have never had jam.

Quote: ChesterDog

It wasn't taught in my high school either. I'm from the U.S., but the original poster is from another country as revealed by his calling math "maths." And I bet most countries have higher math education standards than the United States.

Ha... definitely. This is beyond any high school math I had, as well as many college courses.

Here is a high school math problem I did not get then and do not get now. Algebra II word problem. It basically stated some kid had say $1.65 in a mix of nickels and dimes. It asked how many nickels and how many dimes.

My reaction was and is that it is impossible to say. A limited number of possibilities to be sure, but to limit it to one?

I forget if he used two variables or one, but I do remember it somehow it multiplied the variables by .05 and .10 and added to the total. I do remember there was one and only one answer.

Anyone else remember this problem? FWIW the only real thing I got out of the class was understanding why they had "concentric" shapes and how a parabola behaves and why we look for it in stocks, etc. The rest of the class was many hours of my life I will never get back.

Quote: AZDuffman

Here is a high school math problem I did not get then and do not get now. Algebra II word problem. It basically stated some kid had say $1.65 in a mix of nickels and dimes. It asked how many nickels and how many dimes...

I like that kind of algebra word problem.

Your example, might be: A kid has $1.65 in a mix of nickels and dimes. If the number of his dimes is three more than the number of his nickels, how many of each does he have?

The problem gives two pieces of information, and there are two unknowns.

The unknowns:
Let n = the number of nickels
Let d = the number of dimes

The two pieces of information expressed as equations:
5n + 10d = 165 (This means the money in nickels plus the money in dimes is the total money.)
d = 3+n (This means the number of dimes is three more than the number of nickels.)

Algebraic manipulations:
5n + 10d = 165
5n + 10(3+n) = 165
5n + 30 + 10n = 165
15n = 165-30
15n = 135
n = 9 nickels
d = 3 + n =3 + 9 = 12 dimes

Check the answer:
12 = 3 + 9 check
12(0.10)+9(0.05)=1.20 + 0.45 = $1.65 check

Quote: ChesterDog

I like that kind of algebra word problem.

Your example, might be: A kid has $1.65 in a mix of nickels and dimes. If the number of his dimes is three more than the number of his nickels, how many of each does he have?

The problem gives two pieces of information, and there are two unknowns.

The unknowns:
Let n = the number of nickels
Let d = the number of dimes

The two pieces of information expressed as equations:
5n + 10d = 165 (This means the money in nickels plus the money in dimes is the total money.)
d = 3+n (This means the number of dimes is three more than the number of nickels.)

Algebraic manipulations:
5n + 10d = 165
5n + 10(3+n) = 165
5n + 30 + 10n = 165
15n = 165-30
15n = 135
n = 9 nickels
d = 3 + n =3 + 9 = 12 dimes

Check the answer:
12 = 3 + 9 check
12(0.10)+9(0.05)=1.20 + 0.45 = $1.65 check

I hear ya, maybe I just do not remember the "extra" piece of information. I just remember them saying "a mix." Your problem I get in a second, makes sense. For me the old one is likely lost to time.