[Blindmath] Standard Deviation Question

Sun Nov 2 14:34:27 UTC 2014

To build on Joseph's explanation:
The question said faster than 95% of the entire field. 
In stats, a probability density function, the frequency, frequency density are plotted on the x, y axes respectively. The area between the function's plot & x axis is obviously the total probability, which over the entire possibility space is equal to 1.
Now for any situation, the frequency always increasses over the x axis from left to right.  So in a normalised distribution, z corresponds to frequency values. Since frequency refers to time in your question, Int z(-infinity, +infinity)=1 for the entire possibility space, which is the entire time scale for all your swimmers. Since time, & therefore the corresponding z always increase from left to right, Int z(x), +infinity)=0.95 where x is the answer in question. Therefore, since there would be 5% chance of someone having a faster time than x, Int(-infinity, z(x))=0.05 . 
Now you were very close to your answer. You had found z such that  Int(-infinity, z)=0.95 . Since a normal distribution is symmetrical around 0, Int(-infinity, -z)=0.05 by law of symmetry. 
Given that z(x) = (x-mu)/sigma, plugin the values of z (which should be negative in this case), mu & sigma & you'll get the value of x. :) 

HTH
Muzz

Sent from my iPhone

> On 2 Nov 2014, at 00:26, Joseph Lee via Blindmath <blindmath at nfbnet.org> wrote:
> 
> Hi,
> Hmmm, I think I know what went wrong: a lot of times, statistics requires
> that you look closely at what is being asked. Statistics, like any other
> mathematics disciplines, requires careful attention to detail and wording.
> In this case, the question asked about "faster time", which in this case
> means earlier times. Do not worry about Z-scores now and your brain being
> "fried" - it might be wrong impression you've got somehow that "faster time"
> meant later times.
> Cheers,
> Joseph
> 
> -----Original Message-----
> From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of Sarah
> Jevnikar via Blindmath
> Sent: Saturday, November 1, 2014 5:15 PM
> To: 'Blind Math list for those interested in mathematics'
> Subject: Re: [Blindmath] Standard Deviation Question
> 
> Thank you Marana and Joseph! I did some looking up after writing initially,
> and figured out the following.
> Standardization has to happen. Except that I found the Z-Score for a
> probability of 0.95, which is all but the rightmost tail of the
> distribution. This gives a Z-score of 1.645 and a swim time of 5.187,
> neither of which make sense. I can't think of the intuition as to why I'd be
> looking at the left 5% tail though; my brain must be fried. Any thoughts on
> this? Google's not making it clear and I don't have my stats books handy.
> 
> -----Original Message-----
> From: Ramana Polavarapu [mailto:sriramana at gmail.com]
> Sent: November-01-14 5:56 PM
> To: Joseph Lee; Blind Math list for those interested in mathematics
> Cc: Sarah Jevnikar
> Subject: Re: [Blindmath] Standard Deviation Question
> 
> Obviously, there are better statisticians than I am on this group.
> Joseph has answered your question.  The main thing is to figure out the area
> under the curve equal to 90%.  Here is my thinking:
> 1. She is better than 90%.
> 2. If she hits the mean (4.2), she is better than 50%.
> 3. We need to get rid of 40% below (left of) mean.
> 4. By looking at the tables, you can figure out 1.28 standard deviations
> equal to 40%.
> 5. The answer should be 4.2 - 1.28 times .6.
> I have looked at the following web site to get this answer:
> http://www.dummies.com/how-to/content/how-to-find-a-percentile-for-a-normal-
> distribution.html
> Look at the fish example mentioned there.
> 
> Thank you.
> 
> Regards,
> 
> Ramana
> 
>> On 11/2/14, Joseph Lee via Blindmath <blindmath at nfbnet.org> wrote:
>> Hi,
>> I think you meant 4.2, not 4.52. If it was 4.52, then two standard 
>> deviations must be 3.32-5.72. Also, faster than 95% implies that 
>> Sally's time should be shorter (3.0-3.32). In reality, Sally's time is:
>> 4.52 (or 4.2) - (sigma of 1.95 * 0.6).
>> As a normal distribution is shaped like a bell (really a cone with a 
>> smooth U at the top) with an imaginary line running down the center of 
>> it, a standard deviation measures how data is spread out on the center 
>> of the bell. Thus, one standard deviation will denote data range 
>> between top 16% to bottom 16%, two standard deviation measures from 
>> top 2.5% to bottom 2.5% and so on (what I mean by "top" and "bottom"
>> are really left and right sides; so half of a data that lies within 
>> one standard deviation will be on the immediate left of the center 
>> line). The concept of normal distribution and the bell curve shows up 
>> again when a student studies about inference and standard error.
>> Cheers,
>> Joseph
>> 
>> -----Original Message-----
>> From: Blindmath [mailto:blindmath-bounces at nfbnet.org] On Behalf Of 
>> Sarah Jevnikar via Blindmath
>> Sent: Saturday, November 1, 2014 2:05 PM
>> To: 'Blind Math list for those interested in mathematics'
>> Subject: [Blindmath] Standard Deviation Question
>> 
>> Hi all,
>> I'm helping a friend out with stats homework, but I can't remember the 
>> specifics of a normal distribution and don't have a tactile diagram handy.
>> I'm wondering if someone could help me fill in the gaps.
>> 
>> The question reads:
>> "The average swim time for a 200m race was 4.52min with a standard 
>> deviation of .60min. Sally swam faster than 95% of her competitors in 
>> the race. What was Sally's race time?"
>> 
>> My thinking was as follows:
>> The standard deviation of 0.6 min and the six sigma rule that states 
>> 68% of data is in the first sd, 95 in the second, and 99 in the 3rd,
> should apply.
>> This would mean that 68% of swimmers would have times between 3.6 and 4.8.
>> Then 95% of swimmers should have times between 3.0 and 5.4. If sally 
>> is faster than 95% of swimmers, would this make her time 5.4? I'm 
>> thinking not but I'm not sure what I'm missing.
>> Thank you as always for your help,
>> Sarah
>> 
>> 
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