This is not hw, just a couple problems in lecture I didn't understand

College Life
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<p>1.(More detail): For a group of women, the 25th percentile of height is 62.2 inches and the 75th percentile is 65.8 inches. The histogram follows the normal curve. Find the 90th percentile of the height distribution.</p>

<li>(Different problem) For another problem, the following is applicable:</li>
</ol>

<p>SAT: Mean 550, SD 80
GPA: Mean 2.6, SD 0.6, r=0.4</p>

<p>For those with ___ GPA, 50% have SAT score at least 600.</p>

<p>Can someone clearly explain to me how to solve these two questions. Thanks.</p>

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<p>For problem #1, you have enough information to find the standard deviation. I don’t remember the formula off the top of my head, but I’m sure you have it in your notes or textbook. Once you have the SD, you can apply a normal model, find the z-score of the 90th percentile (from a chart), do a little algebra, and voila.</p>

<p>Mmmmm, I’m assuming that r in the second problem deals with the level of independence between the two variables, which is something that we haven’t covered yet. Sorry.</p>

<p>I don’t have my book or notes on me, and Stats isn’t my best subject, but I hope I’ve at least been a little helpful.</p>

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<p>Thanks, I am now able to do the first problem. The second is still troublesome though. The r is the correlation coefficient.</p>

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<p>I believe this is the way to solve the 2nd problem:</p>

<p>You need to solve the equation</p>

<p>Average SAT + ((x - avg GPA)/SD of GPA)(r)(SD of SAT) = Estimated SAT.</p>

<p>This is using the regression method of estimating one variable from another.</p>

<p>The equation becomes 550 + ((x - 2.6)/0.6)(0.4)(80) = 600.</p>

<p>Solving for x gives you 3.5375. If this is a homoscedastic distribution then 3.5375 is the GPA where 600 is the average SAT score for all of the people with that GPA. The new SD for all the 600 scorers would be (1 - r)^(0.5).</p>

<p>Is your class Stat 2, 20 or 21?</p>

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<p>Thanks for the help, I’d appreciate it if someone could answer one more question. The following is applicable:</p>

<p>Height: Mean 68 inches, SD 2.7 inches
Forearm length: Mean 18 inches, SD 1 inch, r=0.8</p>

<p>Of the men who are 68 inches tall, what percentage have forearms which are 18 inches long, to the nearest inch?</p>

<p>If I use the regression method to find the new average for the forearm length, I get zero. What am I doing wrong? By the way, I am in Stat 20 and I am an econ major.</p>

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<p>bump…</p>