A manufacturing process fills soft drink bottles with a population mean equal to 12.074 ounces and a population standard deviation equal to 0.046 ounces. a. What percentage of the bottles will have less than 12.010 ounces?  b. What percentage of the bottles will have more than 12.172 ounces? 

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We are given that the population mean mu=12.074, with a population standard deviation sigma=.046. We are asked to find the percentage of the population with the following properties:

(a) Find P(x<12.010):

We convert the data point by "normalizing": z=(x-mu)/sigma, so z=(12.010-12.074)/.046 or z is approximately -1.39. (Note that this is reasonable; the z score indicates how many standard deviations a score is from the mean and in which direction. 12.010 is less than 12.074 so z<0, and it is almost 1 1/2 standard deviations from the mean.)

Now P(x<12.010)=P(z<-1.39) (Assuming that the underlying population is approximately normal.)

Consulting a standard normal table, we find that the area of the curve to the left of -1.39 (or equivalently the probability of a random z being less than -1.39) is .0822.

So P(x<12.010 is approximately 8.22%

(b) P(x>12.172): convert to a z score; z=(12.172-12.074)/.046=2.13

Then P(x>12.172)=P(z>2.13)

If we are using a table we must be careful, as most tables give the area to the left; here we want the area to the right so we find 1-P(z<2.13)=1-.9834=.0166

So P(x>12.172)=1.66%

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