Normal Distribution

Question

The labels on boxes of Cheesy Poofs say that the box contains 16 ounces of Cheesy Poofs, but that doesn’t guarantee that each box contains exactly 16.00 ounces. The machine that fills the boxes doesn’t always put out the exact same amount. The amount in each box is Normally distributed, and has a standard deviation is .40 ounces. To help assure that most boxes contain as much or more than the label says, the machine is set so that the mean will be 16.28 ounces. 

g) Half of the boxes of Cheesy Poofs produced which are labeled 16 ounces, actually contain more than 

________________ounces

h) 14% of the boxes of Cheesy Poofs produced which are labeled 16 ounces, actually contain more than 

_____________ounces. 

Answer 1

Cheesy Poofs. I love it.

So, let's name a random variable X to represent the number of ounces of Cheesy Poofs being poured into a box. When I write P(X>x) = ___, I mean that the probability of the amount of Poofs being poured into a box being greater than x ounces is...equal to something.

So, let's do some inverse operations. Your two-part question has percentages which need to turn back into raw measurements of amount of Poofs poured in a bag. On a TI-83 (and should be on TI-84 as well), this is done by the invNorm() function. Access it by going to DISTR (which is 2nd VARS) --> 3: invNorm( .

If you're on the TI-83 (which does not prompt for values to plug in), follow this order:

invNorm(probability, mean, standard_deviation).

For the first part, we need to find the amount of Poofs in half of the boxes. The "more than" language indicates that we need to study a part of our normal distribution, our curve, that covers a z-quartile of interest and beyond, not from the left-up (that works for a "less than" question).

Type: invNorm(1-0.5, 16.28, 0.4). This gives the upper 50% of boxes. This answers the question of

P(X > ??? ) = 0.5. We get 16.28.

50% of boxes of Cheesy Poofs produced actually contain more than 16.28 ounces.

For the second part, we're looking at P(X = ???) = 0.14. I need the highest 14% of boxes, since they will be the heaviest.

Type: invNorm(1-0.14, 16.28, 0.4). This gives the upper 14% of boxes. We get 16.71.

14% of boxes of Cheesy Poofs produced actually contain more than 16.71 ounces.

Please let me know if I can clarify anything, or if you need a way to solve this problem without technology.