# Central Limit Theorem ???

#### jena2007

##### New Member
Hi,

I'm having trouble understanding this problem. Our professor gave the answer to study with, but I can't understand why the probabilities would be set up as
P(4590 > X > 4610) rather than P(4590 < X < 4610). Here is the question below:

Averages of several measurements are less variable than individual measurements. The true mass of the whiskey sample in Exercise 5. 28 is 4.6 grams, or 4600 milligrams (mg). Antonio’s measurements have the normal distribution with mean 4600 mg and standard deviation 10 mg. In this case, the mean of his 3 measurements also has a normal distribution.

b. What is the probability that Antonio misses the true mass by more than 10 mg in either direction if he makes one measurement?
X~N( 4600, 10) so, P(4590 > X > 4610)

z 4590 = (x - µ)/σ = (4590-4600)/10 = -1

z 4610 = (x - µ)/σ = (4610-4590)/10 = 1

P (-1 > z >1) = .1587 + (1-.8413) = .1587 + .1587 = .3174

c. What is the probability that the mean of three independent measurements misses the true mass by more than 10 mg?
0~N( 4600, 5.77) so, P(4590 > 0 > 4610)

z 4590 = x - µ/σ = 4590-4600/5.77 = -1.73

z 4610 = x - µ/σ = 4610-4590/5.77 = 1.73

P ( -1.73 > z >1.73) = .0418 + (1-.9582) = .0418 + .0418 = .0836