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%