<p>In a certain city 6% of teenagers are married, 25% of married teenagers have children, and 15% of unmarried teenagers have children. If a teenager has a child, what is the probability that the teenager is not married?</p>
<p>a) .156
b) .2
c) .5
d) .904
e) .940</p>
<p>(.94<em>.15)/(.94</em>15+.25*.06)</p>
<p>d) .904</p>
<p>In post #74, how do you define the parameter?</p>
<p>invnorm is for z scores</p>
<p>Is the free response section done in pencil or pen?</p>
<p>I’m so screwed! :D</p>
<p>Probably pen. Shouldn’t make a difference though.</p>
<p>Actually, almost certainly pen.</p>
<p>How do you do this problem?</p>
<p>Lengths of individual shellfish in a population of 10,000 are normally distributed w/ mean 10 cm, std dev of 0.2 cm. Which is the shortest interval that contains 4,000 shellfish lengths?</p>
<p>Answer : 9.895 cm to 10.105 cm</p>
<p>40% of the population of a normal distribution is found between a certain set of z scores. To use invNorm, you would input 70%, since its one tail (and there is 30% on each side of the 40% interval you are looking at.) You get a z-score of +/-.5244. When multiplied by .2, you get an x value of +/-.1048, rounded to +/-.105. These results in the interval found int he answer you included in your post.</p>
<p><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></p>
<p>need help on #1. sad i know.</p>
<p>k I’m just gonna go to bed after this problem… help…
The mean and standard deviation of the x values are x =5 , sx =4 . y=10, sy=10.
What is the least squares regression line?</p>
<p>How do you guys calculate coefficient correlation?</p>
<p>@miracle </p>
<p>count the number of dots up to 85 miles.
then you realize you the dot on 85 miles is the 20th dot :which means its at 20/28 percentile, 71%… B.</p>
<p>@ <a href=“http://www.kent.k12.oh.us/~ke_bmccom...sed%20exam.PDF[/url]”>http://www.kent.k12.oh.us/~ke_bmccom...sed%20exam.PDF</a></p>
<p>How do you find the standard deviation in #16?</p>
<p>@ eqweewee</p>
<p>so expected total number of orange candies is just 0.1(25) + 0.45(25) which gives you 13.75. Standard deviation is a little tricky. You know that variance = np(1-p). In this case, you have to add up the variances for dynamite mint candies and holiday mint candies AND THEN square root your final answer (this is how you combine independent random variables). Doing the calculation, you get (25 x 0.1 x 0.9) + (25 x 0.45 x 0.55) = 8.4375 and then square root 8.4375 to get 2.905.</p>
<p>and @ howon,</p>
<p>go to the formula sheet for descriptive statistics. notice the formulas b0 = y bar - b1 (x bar), b1 = r (sy / sx), and y hat = b0 + b1x. I think you might need to be given b0 to simply plug in the values into these equations and solve for r (correlation coefficient) and your least squares regression line (the y hat equation).</p>
<p>Can someone please tell me how to calculate quartiles?
Eg. for-
1 6.6
2 8.2
3 8.8
4 9.5
5 10.7
6 10.8
7 10.9
8 13.2
9 13.3
10 14.3 </p>
<p>How can we get Q1: 8.975; Median: 10.75; Q3: 12.630</p>
<p>i took it today. its not too bad dont worry about it so much</p>
<p>^ Any trick questions that I could prep for?</p>
<p>Can someone PLEASE PLEASE explain how to calculate quartiles?
Eg. for-
1 6.6
2 8.2
3 8.8
4 9.5
5 10.7
6 10.8
7 10.9
8 13.2
9 13.3
10 14.3</p>
<p>How can we get Q1: 8.975; Median: 10.75; Q3: 12.630</p>
<p>do you know how to find the median?
just find the median first, then find the median of the lower half (to get Q1) and the median of the upper half (to get Q3)</p>
<p>i think…</p>
<p>Quartiles are the median of the lower and upper halves.</p>
<p>irunoninsulin got it in before me. He/She is correct.</p>
<p>I’m actually just about to leave for this exam. Best of luck to all of you!</p>