<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Can somebody please explain to me how to do problems 7, 11, 14, and 16 of the multiple choice?</p>
<p>please help??</p>
<p>also, from the same link, can you explain #6 of the free response? i don’t think i even know what it is asking</p>
<p><em>bump</em></p>
<p>i really need help please guys. i thought i was in the solid 5 range but i’m missing seemingly easy questions like these please help!</p>
<ol>
<li>Each person has a mean weight of 190 and SD of 10. </li>
</ol>
<p>My mistake was multiplying 4 to each of the statistics. However, that gives me a wrong answer because the context of the problem states, “Suppose the distribution of adult males who rent boats…,” which means that we assume that the given data are parameters (statistics of a population). </p>
<p>Therefore, 800/4=200 and since 4 is the sample size, SD =10/sq4=5. So, normalcdf(200,10^99,190,5)=.02275. A is the answer.</p>
<p>Or another variation. You can multiply 4 to each of the parameter, but since there’s still the sample size, SD=40/sq.4=20. So, normalcdf(800,10^99,760,20) </p>
<ol>
<li>You can solve this similarly to the second variation for number 7.</li>
</ol>
<p>4(SD)x10 (number of bars within the box)=40. Since there is a sample, you have to find the SD for the sample size: 40/sq10. From here, you have to combine the SDs of the weights of the bars and the box, so sq((40/sq10)^2+8^2)=sq224. </p>
<ol>
<li>A) y=-5.0+3.0(5)=10 good
B)y=3(5)=15 not good
C)y=5.0+2.5(5)=17.5 not good
D)y=8.5+.3(5)=10 good
E)y=10.0+.4(5)=12 Not good</li>
</ol>
<p>So, you have the choice between A and D. The last step is this:</p>
<p>b=rs(y)/s(x)=r(10/4)=5r/2</p>
<p>So, b has to be between -5/2 and 5/2 since the range of r is -1 and 1. The answer is therefore D since .3 is within the range</p>
<ol>
<li>(d) Because George and Michelle used the same questionnaire, their results will
generalize to the combined population of their friends. </li>
</ol>
<p>This reasoning is false because their results are being compared by two different population and combining the population will not tell you the result. I’m shaky on this reasoning though.</p>
<p>Thank you so much!! I was also multiplying each one by 4 for #7 and made the same error on #11. I never learned #14’s method. #16 is still a little confusing, but your choice (the correct answer) makes more sense to be false than any of the others. Thanks!</p>