<p>oh cool. and whats a CLT? lol</p>
<p>umm i said it was from a normal population because the number was like 1.03…meaning that the two things in the fraction (idr what they were) were pretty close together…so it wasnt that skewed or something</p>
<p>And do you guys remember 6b? The choices were weird, it said (large, small, close to 1)…I put larger than 1…but many people are telling me its just large?</p>
<p>CLT = Central Limit Theorem.
States that for large enough ‘n’, the shape of the sample mean (x-bar) will be normal regardless of the shape of the true population mean.</p>
<p>how did you guys assign two groups of equal size? i said to use a random number table (0-4 for one, 5-9 for the other group), and if the two resulting groups are different sizes, randomly select however many people you need and move them to even out the groups.</p>
<p>but i’m sure there’s some better way.</p>
<p>this is what i did (dont know if its the best way): assign each child in the group of 24 a number from 01-24. Reading pairs of digits from a random number table, and throwing out 00 or anything above 24, assign the first 12 numbers called to one group, and the rest go in the other group.</p>
<p>I’ll just do all my work out for #2.</p>
<p>A.) first find the z-score for the 70th percentile (z=.5244)
M=125, SD=6.5</p>
<pre><code> .5244 = (x-125)/6.5
x-125=3.4086
x=128.4086
</code></pre>
<p>B.) Binomial situation where p=.3 and the number of trials is 5.</p>
<p>summation(from 2 to 5) 5 C X (.3)^X (.7)^(5-x) = .47187</p>
<p>C.) Sampling distribution were M=125 and SD=6.5/√5</p>
<pre><code> Z=(130-125)/(6.5/√5)
Z=1.72005
look up probability (p=.0427)
P(mean stopping distance at least 130)=.0427
</code></pre>
<p>Got the same answers NYEM. Hopefully we’re both correct!</p>
<p>same here! woo! i totally thought i was doing the wrong thing for part C</p>
<p>funny thing is, I merely just skimmed through FRQ’s online from the CB website and remembered howta do part c as soon as I saw it. Haha.</p>
<p>Yep those are right. Its a z-test, not t.</p>
<p>For 6b, the one that said large, small, or close to 1, isnt it just Large??</p>
<p>I got confused to because it has to be a certain largeness…</p>
<p>Gahh, why is it a z test and not a t-test???</p>
<p>because the sample is small? idk</p>
<p>is the probability still the same though?</p>
<p>I’m pretty sure it was a t-interval. Even though n was large enough, the two population SD’s were not known and we used the sample means as well.</p>
<p>Not a t-test. It’s a two sample t interval.</p>
<p>oh oops, nvm</p>
<p>^If it is a t-test, then your answer is wrong, as it would be t(4)>1.72, not z>1.72. The probablilites WOULD NOT be the same. I actually think that you are right (I thought it was a t-test) because the SD of the sample of 5 compes from the sample SD, which is the population SD in that case.</p>
<p>Btw, so whats the answer for 6b? Large? Anyone?
Thanks</p>
<p>for 6b, i said the statistic was skewed right because it greater than 1. If distribution is skewed to the right, then the mean will always be larger than the median</p>
<p>I wasn’t sure what they meant by “large” so I also put greater than one.</p>
<p>For the last part on number 6, did you just flip the fraction they gave you?
(median/mean)
Now if it is LESS than 1, the data skewed to the right
Would that be a valid answer or is it too simple?</p>
<p>No. The mean would not be included in the five number summary; they were looking for a statistic that could judge skewness based solely off of the five numbers they provided.</p>
<p>oh crappp, i didn’t even look at that chart : (</p>
<p>wat score would i get for that problem if that part is the only incorrect answer?</p>