Central Limit Theorem ???

#1
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

Answer: .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

Answer: .0836

Thank You:)
 

JohnM

TS Contributor
#2
you may have a point - why don't you ask him about it?

It looks like he computed the probabilities for "within" 10mg when the question is looking for "outside" of 10mg