<p>just one more</p>
<p>Random variable X is normally distributed with mean 10 and standard deviation 3 and random variable y is normally distributed with mean 9 and standard deviation 4. if x and y are independent, which of the following describes the distribution of Y-X</p>
<p>B. Normal with mean -1 and standard deviation 1
C. normal with mean -1 and standard deviation 5</p>
<p>So if i put it into my calculator (X= L1 = 4,7,10,13,16; Y=L2 = 1,5,9,13,17) and subtract L2-L1 i get (-3,-2,-1,0,1), which has a sd of 1. However if i use the formula (root(3^2 +4^2)) i get sd as 5. Which ones the right answer?</p>
<p>It’s gotta be SD = 5. Standard deviation is trigonal, which means that it depends on the squares of the two constituent standard deviations, and not just the difference between them. So like you said, it’s sqrt(9+16)</p>
<p>@chocolatecricket
The problem with your lists L1 and L2 is that they don’t actually have a standard deviation of 3 or 4, respectively, just common differences of 3 and 4 and the appropriate means.</p>
<p>Always use the second (square-root of the sum of the variances) method because it is guaranteed correct and is much easier and less risky.</p>
<p>I have a question from the 2002</p>
<ol>
<li>As lab partners, Sally and Betty collected data for a significance test. Both calculated the same z-test statistic,
but Sally found the results were significant at the a = 0.05 level while Betty found that the results were not.
When checking their results, the women found that the only difference in their work was that Sally had used </li>
</ol>
<p>I think it is B, since the area to the left of -1.690 is about .0455, while twice this is more than .05 and thus not significant. However, it also seems like D and E could be correct if you do the area to the right…</p>
<p>thats weird…the thing i have says its A.
I also thought it was between B and D</p>
<p>Oh sorry, I misread. SALLY found them significant while Betty did not. In fact, the only one that is significant for a 2-tailed test at a = 0.05 is -1.980. If Betty used a right-tailed test, the significant p-values would all be the areas beyond z scores of 1.64 or greater, and
-1.980 is clearly not in the rejection region for this.</p>
<p>OH wow i made the same mistake LOL</p>
<p>can somebody explain this to me?
The distribution of the diameters of a particular variety of oranges is approximately normal with a standard deviation of .3 inch. How does the diameter of an orange at the 67th percentile compare with the mean diameter?</p>
<p>a.) .201 below the mean
b.) .132 below the mean
c.) .132 above the mean
d.) .201 above the mean
e.) .440 above the mean</p>
<p>^^ It’s C. Do invNorm(.67) to find the value of the z-score (or look it up in the table). You get a z score of about .44. Multiply by .3 to get .132; this is .132 inches above the mean (note: do not actually round until the very last step).</p>
<p>Also, if you couldn’t do that, you had better start studying, no offense just saying.</p>
<p>@commodore
The answer should be C
The way you do it is first invnorm(.67). This gets you the z score of the diameter
Then you multiply .3 times this number and u get .132. Its above the mean because its the 68th percentile, which is above 50 (the middle value)</p>
<p>What about this one? Its really hard 
<a href=“http://www.kent.k12.oh.us/~ke_bmccombs/apstatsl/Exam%20review/2008%20released%20exam.PDF[/url]”>http://www.kent.k12.oh.us/~ke_bmccombs/apstatsl/Exam%20review/2008%20released%20exam.PDF</a> number 40</p>
<p>look at this site: [Central</a> Limit Theorem (parabola)](<a href=“statisticalengineering.com”>statisticalengineering.com)</p>
<p>I recommend memorizing what all of them look like.</p>