1. ## Probability Q Again..

Hi...I have another question><

The weight of the contents of a can of Campbell's minestrone soup can be modelled by a normal distribution with a mean of 439.9 grams and a standard deviation of 3.2 grams. A random sample of 15 cans is selected for quality control testing. Determine the probability that (4 dec pl)
1. a randomly selected can weighs more than 437.5 grams
2. the sample mean of the cans taken for quality control testing will NOT be within the target range of 438.0 to 442.0 grams.

thx

2. 1.
From the given information we can find the z-score for this probability:

Z = (Score - Mean) / Standard Deviation
= (437.5 - 439.9) / 3.2
= -0.75

Going to a standard z-score table and locating -0.75 we are given a probability of 0.2266, or 22.66%.

The z-score will give the probability of LESS THEN OR EQUAL TO the score in question. Since we're looking for the probability that the chosen can is AT MORE THAN 437.5g, we look at the probability of the can NOT being at least 437.5g.
In numeric terms, this equals

= 1 - 0.2266
= 77.34%
So there is a 77.34% chance of the selected can weighing MORE than 437.5g
(Sorry, the z-table I use only goes to 2 decimal places)

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2.

For the next section, on a normal distribution curve we're interested in the section between the z-score corresponding with 438.0g and 442.0 grams.

So, first off we need to find the z-score for each of these weights. This is calculated the same way as in the first section.
Z = (Score - Mean) / Standard Deviation
Z(438)= (438 - 439.9) / 3.2
Z(438)=-0.59375
~-0.59

Z(442)= (442 - 439.9) / 3.2
Z(442) = 0.65625
~0.66

Going back to a z-score table we find the corresponding probabilities.

-0.59 = 0.2776

0.66 = 0.7454

Since each of these gives the probability of less than or equal to, we can find the probability between the two values by subtracting the smaller one from the larger one.

0.7454 - 0.2776 = 0.4678 - However this gives the probability of the sample being within the selected range. The complimenty event (NOT being within the selected range) is found by subtracting this probability from the total probability (1).

1 - 0.4678 = 0.5322 = 53.22%
There's a 53.22% chance the selected sample will NOT be within the selected range.

Oh dear, sorry I just noticed question 2 might not be what you're after. That is the probability that any can chosen at random will not lie withing this range. I can't quite get my head around the 'sample mean of the cans' part. Sounds to me like you need to somehow involve the standard deviation to find if the mean of the random 15 cans lies within this range. It's almost 2am here in Australia so I'll have another think about it when I'm more mentally 'with it' and post again if I can think of anything useful.