An interesting question...

[/0]

1994 MM:

Wizzi Cherry Drink is sold in 500 mL bottles. The company determines that the volume of drink in each bottle is normally distributed with mean 498 mL and standard deviation 2.5 mL.

The probability that a bottle selected at random will contain more than 500 mL is equal to

A. 0.1056
B. 0.2119
C. 0.5000
D. 0.7881
E. 0.8944

What would you do?

[Collin Li]

Let X be the volume of drink in each bottle.







You need a calculator. The work I did isn't actually necessary, because the calculator does the Z-score calculation for you. The relevant function is normalcdf(), found in the TI-83 by going [2nd] [DISTR].

Syntax: normalcdf(lower bound, upper bound, mean, standard deviation)

However, we do know that the answer must be less than 0.5, because we are looking for the area under the curve for z = 0.8 (z = 0 is the centre of the graph). (This will make no sense if you have not done normal distributions before)

For this question, as an example: normalcdf(500, 1e99, 498, 2.5). "1e99" is 1 followed by 99 zeroes, so that is supposed to represent positive infinity. I got 0.2119, the answer seems to be B.

[/0]

Thanks coblin! , I'm a bit weak with the calculator.

But don't you think it's weird that they're asking for the probability that the bottle is filled beyond its limits? Since its pretty hard to compress water, it's not likely it would go past 500 mL, so if I didn't know better I would just put F. 0. LOL

[Collin Li]

I think it meant that it's sold in bottles marked 500mL. There's always room to bring it over, and there is pressure in the bottle

Don't worry about that too much, otherwise they'd make it clear (and talk more about the bottle structure to allude to the fact that it's not physically possible -- but then it would no longer obey normal distribution, so it's just not viable to interpret it that way).