Normal Distribution Question Check!

ShaShaBear

Okay, so I plan to be frequenting this board quite a bit as it is my go-to for checking my progress on learning how to solve problems for my statistics exam in two weeks (gets me into final year of college and I have literally no clue!)

This is the question I have been working on:
A normally distributed random variable has mean 40 and standard deviation 4. A sample of 4 observations is generated from this, and their mean xbar is calculated. What is the probability that xbar is:
(i) larger than 43
(ii) smaller than 39
(iii) between 39 and 43

Using a mixture of stattrek and mathsisfun website tutorials, I managed to come up with this:
(i) Xbar~Normal(mu=40, sigma=4/sqrt4) => Xbar~Normal(mu=40, sigma=2)
Z=(X-mu/sigma)
P(Xbar>43) => P(Z>[43-40/2]) => P(Z>1.5)
Z-table for 1.5, .00 is 0.0668, or 6.68%
(ii) P(Xbar<39) => P(Z<[39-40/2]) => P(Z<-0.5)
Z-table for 0.5, .00 is 0.3085, or 30.85%
(iii) P(39<Xbar<43) => P(-0.5<Xbar<1.5) => 0.3085-0.0668 = 0.2417 or
24.17%

Have I done this right? I cant find a calculator online to verify my answers! Also, if I have done this right, can I assume any question phrased like this will be done in the same way, or are there variations?

MathsManiac

I haven't checked your table-lookups, but you're method seems correct.

ShaShaBear

The table I used is an interactive one on mathsisfun.com
I can use the proper one, but since all I do is slide it over to the Z number, it tells me the answer so I don't make mistakes

If I have the method right I guess that's the main thing, thanks!

hivizman

MathsManiac wrote: »

I haven't checked your table-lookups, but you're method seems correct.

Yes, the method is correct, but the answer to part (iii) doesn't look right. The three cases (i) larger than 43, (ii) smaller than 39, and (iii) between 39 and 43 are mutually exclusive and exhaustive (that is, any observation must fall into one and only one of these three categories), so the three probabilities must sum to 1.

In part (iii), I think that the OP is correct down to expressing the probability as P(-0.5<Xbar<1.5), but this is equal to P(Xbar<1.5) - P(Xbar<-0.5), which is not what the OP has [the OP has P(Xbar<-0.5) - P(Xbar>1.5)].

I believe that the answer is (1 - 0.0668) - 0.3085 = 0.9332 - 0.3085 = 0.6247, or 62.47%.

Note that 62.47% + 6.68% + 30.85% = 100.00% = 1.

ShaShaBear

In a moment of silliness, I switched them around so that the smaller number would be taken from the bigger number

As I said, very, VERY bad at stats and I will likely be posting once a day for the next two weeks trying to get answers to the supposed obvious. Thanks so much guys, that's two questions out of the way, got a few different types of probability to work on today - no doubt I'll be back!

MathsManiac

Sorry I didn't spot that - I just saw that the OP was doing part (iii) by combining parts (i) and (ii) and didn't look carefully enough at the detail!