<p>Can someone help me with this prob and tell me the exact step sequence on the calculator?</p>
<p>the question is 2.31 in the Practice of Statistics by yates, moore, and starnes.</p>
<p>Thanks</p>
<p>Next time put the question up before you go and ask it here. Not all of us have the book, but you're lucky I do :P</p>
<p>I'll be assuming you have a TI-83 for this
Go to 2nd, then distr , then go to normalcdf( than input your lower bound, than upper bound, then mean, then standard deviation, and there you go, so that's 4 values you need to put in</p>
<p>Oh and also, if one of your bounds in infinity, use 10^-99 or 10^99</p>
<p>are you sure? i thought it was normalcdf(lowerbound, Z-score)
normal cdf (z-score, upperbound)</p>
<p>take a look at this link,</p>
<p>Example. What percentile is 1150 on SAT, if mean SAT is 1100 and standard deviation is 100? (those are made-up parameters)</p>
<ol>
<li>Z = (X - m)/s = (1150-1100)/100 = 0.5</li>
<li>percentile = normcdf(-100,0.5) = .69, or the 69th percentile</li>
</ol>
<p>why do you put in a -100 in "(-100,0.5)" instead of just (100, 0.5)?</p>
<p>The length of human pregnancies from conception to birth varies according to distribution that is approximately normal with mean 266 days and standard deviation 16 days.</p>
<p>A) What percent of pregnancies last less than 240 days (that's about 8 months)?
B) What percent of pregnancies last between 240 and 270 days (roughly between 8 and 9 months)?
C) How long do the longest 20% of pregnancies last?</p>
<p>What are the exact steps for each problem?</p>
<p>-Thanks.</p>
<p>Well, if you're just using the z-scores, there's really no need to use the calculator, you can just use the tables.</p>
<p>The syntax that I have there calculates the area under a normal distribution with those bounfs given a certain mean and standard deviation. There's no need to calculate a z-score for that.</p>
<p>I'm definitely positive about this one. If you type in normalcdf(10^-99,240,266,16), you should get the answer that's found in the back of the book for part a</p>
<p>Note that all the components needed to calculate the z-scores are there. The calculator automatically does it for you if you use that syntax.</p>
<p>any body else?</p>
<p>Straight up use Table A dude. It's pretty simple.</p>
<p>A)
Step 1: z = (240-266)/16
Step 2: Find that Z score in Table A. Find the corresponding answer and change it to percents.</p>
<p>B)
Step 1: z = (270-266)/16
Step 2: Find that Z score in Table A. Find the corresponding answer and change it to percents. Subtract this from the answer to Step A. There's your answer.</p>
<p>C) Since it asks for the 20 LONGEST, you need to find .8000, not .2000
Step 1: Find the z score for .8000 in Table A.
Step 2: Use that z score in the equation z = (x-266)/16.
Step 3: Your answer is all days longer than x and x.</p>
<p>I'd get you the exact answers since I have the textbook on a shelf in my bedroom but I'm too lazy =[</p>
<p>Hope you could decifer my steps lol.</p>
<p>So I got off my butt and got my homework notebook. Here's the answers.</p>
<p>A) The z score is -1.625 and the percent is 5.2% (.0521) So 5.2% of pregnancies last less than 240 days.</p>
<p>B) The z score for 270 is .25 and the percent is 59.8%.<br>
59.8 - 5.2 = 54.6% of the pregnancies last between 240 and 270 days.</p>
<p>C) .8000 is equal to the z score .845. .845 = (x-266)/16. x = 279.52 days.
So the longest 20% of gestation periods are greater than 279 days.</p>
<p>The answers to all odd numbered questions are in the back of the textbook btw. Why did you want to know how to do it on a calculator though? This is a problem where its 10x easier to do it by hand with Table A (which they give you on the AP exam).</p>
<p>Feel free to post any more questions you have about Stats. I like the class so its cool =]</p>
<p>for part C for example, there's this feature called (invnorm), which i don't understand how 2 use</p>
<p>But thanks i understand parts A and B now.</p>
<p>For invNorm, I believe you put in a value between 0 and 1, that corresponds to the area under the normal distribution, and you will get the z-score from it</p>
<p>Can anyone help me with this problem from the book "The Practice of Statistics". Problem 2.32.
When the Standford-Binet IQ test came into use in 1932 it was adjusted so that each age group of children followed roughly the normal distribution with mean = 100 and std deviation = 15. The test is readadjusted from time to time to keep the mean at 100. If present day american children took the 1932 stanford-Binet test, their mean score would be about 120. The reason for the increase in IQ overtime are not known but probably include better childhood nutrition and more experience in taking test.
(a) IQ scores above 130 are often called "very superior". What percent of children had very superior scores in 1932?
(b) If present day children took the 1932 test what percent would have very superior scores? (assume the std deviation 15 does not change.</p>
<p>Anyone know the answer???</p>
<p>Just out of curiosity, isn't 10^-99 = 1/(10^99), which is approximately 0? Why not use 0 if that's what you mean?</p>
<p>On the other hand, if you mean negative infinity, don't you need -10^99?</p>
<p>As far as the new question here:
(a) 130 represents a z-score of 2, meaning that approximately 95% have a score below this, and therefore, that 5% have a score above this.
(b) 130 would represent a z-score of 2/3 if the mean changes to 120, and I don't have a z-score chart handy, so you'll have to take it from here.</p>
<p>yeah thats what i meant, mess ups like this are common with you type too fast and don't stop to think about it</p>
<p>Actually the answer to a is 2.5%. I answered your PM about this question btw with the complete explanation mate =]</p>
<p>The answer to b is 25.5%, check the PM for the work.</p>