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- August 23rd 2012, 09:10 PMch5112help
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- August 23rd 2012, 09:28 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
Start by converting this to a standard normal distribution. The lower quartile is where 25% of the data lies below, so you need to read off where 25% of the data lies below from a standard inverse normal table. Same for the upper quartile where 75% of the data lies below. Once you have the Z values, convert back to your given distribution.

- August 23rd 2012, 09:43 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
Hi Thanks for your help.

On my calculator, I found Z-interval for lower quartile (99.775, 100.23)

for, upper quartile, the z-interval was (99.187, 100.81)

Is this right?? - August 23rd 2012, 09:48 PMMaxJasperRe: help with finding five-number summary when mean, sd and sample size is given
You need to generate 200 real numbers whose mean=100 and sd=10...for example use the following data:

{86.7242,103.095,113.867,100.546,100.32,91.4765,10 2.592,104.713,100.67,113.748,87.5385,92.1847,93.68 7,105.191,109.376,117.19,107.898,97.2232,101.785,8 5.8895,104.347,108.525,89.1039,109.471,92.2035,88. 6707,90.0258,122.608,87.0634,119.748,100.419,82.91 12,94.2178,98.6578,110.744,100.762,93.4528,96.5865 ,103.162,111.058,103.891,114.719,102.908,86.7591,1 10.698,110.139,103.647,95.1995,108.781,105.425,110 .117,99.5495,112.593,95.5003,97.7986,87.7673,99.32 2,96.8874,98.545,107.215,98.2278,80.842,119.057,10 2.44,108.739,85.7724,100.906,94.714,100.761,96.884 2,96.6799,102.845,94.5768,103.694,131.178,111.151, 97.8677,81.6983,80.9117,94.5997,92.3242,102.16,99. 0763,115.843,80.4044,102.292,95.0226,100.146,113.2 42,84.5126,96.8718,91.1545,111.939,94.154,114.719, 96.9888,107.176,114.9,95.7682,110.252,110.461,103. 651,102.557,94.5806,88.8081,90.659,114.598,101.497 ,102.034,106.38,112.795,107.006,91.391,118.129,117 .313,97.4418,78.5687,82.7369,93.2231,119.27,87.115 2,74.5946,90.9401,108.254,91.4419,121.539,99.4256, 83.5403,103.917,109.746,107.337,94.9554,102.363,10 9.493,95.9816,101.473,89.2975,88.6838,121.588,107. 681,99.3644,107.338,85.3285,122.726,87.0816,119.27 1,101.825,83.9203,79.0244,93.4866,105.461,120.249, 94.3469,92.2688,107.272,121.888,105.128,98.2146,10 3.673,84.7568,86.7944,102.236,82.763,100.965,96.07 66,111.454,113.189,111.155,90.9439,99.1924,100.573 ,105.792,93.7059,104.366,100.996,103.198,87.7051,1 11.177,95.2342,107.831,93.4341,108.662,115.518,96. 3542,89.2625,86.7332,97.8433,105.829,85.4472,88.91 34,101.658,99.5822,91.6179,91.3036,108.929,105.385 ,91.1276,98.8022,112.458,90.8502} - August 23rd 2012, 09:48 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
No. You aren't looking for an INTERVAL for a lower or upper quartile, you are looking for a VALUE.

I suggest you read this table, and recall that the normal distribution is symmetric about the mean (which is 0 in the standard distribution). - August 23rd 2012, 09:56 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
so...from the table, it says, for 0.75, it is 0.6745.

Is this the value that i need to obtain??

how about for 0.25?? - August 23rd 2012, 10:01 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
for lower: is it: 93.255

for upper: is it:106.745

after converting it? - August 23rd 2012, 10:03 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
Yes. Now remember that the data is symmetric about the mean. You have where 75% of the data is below (or where 25% of the data is above). How do you think you could reverse this so that you can tell where 25% of the data is below (or where 75% of the data is above)?

- August 23rd 2012, 10:06 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
0.75, it is 0.6745.

so for 0.25, it should be: 1-0.6745 = 0.3255

Then, reversing back, the values are:

for lower: is it: 93.255

for upper: is it:106.745

Is this correct???

Now, I just found Q1, Q3... how can i find minimum, median and maximum?? - August 23rd 2012, 10:06 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
- August 23rd 2012, 10:09 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
Not quite, your VALUE for 25% on the standard normal distribution is the same distance away from the mean as the 75% value, just in the other direction. So the value is -0.6745, but you get the right value for the lower quartile of your set of data.

Now your median should be obvious, it's where 50% of the values lie below. What value is that the same as in the normal distribution?

The maximum value should be obvious, it's where 100% of the values lie below. Can't you read that (or a very close value to it) from your table?

From there you should be able to get the minimum value. - August 23rd 2012, 10:14 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
so for median... the value will be 100 after converting back??

I am not sure of maximum... - August 23rd 2012, 10:15 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
so with maximum, it will be 0.999 right? so the value will be 3.0902

and for median..0.000 - August 23rd 2012, 10:16 PMch5112Re: help with finding five-number summary when mean, sd and sample size is given
value i found for maximum is: 130.90

for median: 100

minimum:69.098

is this right?? - August 23rd 2012, 10:17 PMProve ItRe: help with finding five-number summary when mean, sd and sample size is given
Please be careful with how you word what you are trying to say. If you are trying to say "for the maximum value, we need to look for the value very close to 1 on the inverse standard normal tables", then yes. And then convert that value back.

As for the median, yes, its standard value is 0, which will give 100 when you convert back (there's no need to convert to the standard distribution in this case though). It is well known that for a normal distribution, the mean and median are equal.