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70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 15:18

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70 75 80 85 90 105 105 130 130 130

The list shown consist of the times,in seconds, that it took each of 10 school children to run a distance of 400 meter. If the SD of ten running times is 22.4 seconds,rounded to nearest tenth of second,how many of the 10 running times are more than one SD below the mean of the 10 running times?

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 15:32

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sushma0805 wrote:

70 75 80 85 90 105 105 130 130 130

The list shown consist of the times, in seconds, that it took each of 10 school children to run a distance of 400 meter. If the SD of ten running times is 22.4 seconds, rounded to nearest tenth of second, how many of the 10 running times are more than one SD below the mean of the 10 running times ?

a. one b. two c. three d. four e. five

"How many of the 10 running times are more than one SD below the mean" means how many data points from given 10 are less than mean-1SD.

We are given that SD=22.4, so we should find mean --> mean=100 --> there are only 2 data points below 100-22.4=77.6, namely 70 and 75.

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 20:41

"How many of the 10 running times are more than one SD below the mean" means how many data points from given 10 are less than mean-1SD.

Bunuel, I don't get this - It says how many are more than mean-1SD (one SD below the mean)? Why do you say less? Please explain. Thanks. _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 20:50

How many of the 10 running times are more than one SD below the mean.

So then 1SD below the mean is 77.6. It asks how many of the 10 running times are more than 77.6? I just replaced the one SD below the mean by the value. There are more than 2... What am I missing? _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 21:01

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mainhoon wrote:

How many of the 10 running times are more than one SD below the mean.

So then 1SD below the mean is 77.6. It asks how many of the 10 running times are more than 77.6? I just replaced the one SD below the mean by the value. There are more than 2... What am I missing?

You have a problem with wording.

WRONG reading: How many of the 10 running times are more than one SD below the mean. So more than Mean-SD.

CORRECT reading: How many of the 10 running times are more than one SDbelow the mean. So less than Mean-SD. _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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03 Sep 2010, 21:25

Ok Bunuel, I see what you are saying.. I did not know if everyone will pick it up at one go.. So what do you think the wording would have been if they really wanted to say what I was thinking i.e., the number of running times that are more than 1SD below mean, meaning more than 77.6? _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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10 Sep 2010, 22:03

Bunuel wrote:

sushma0805 wrote:

70 75 80 85 90 105 105 130 130 130

The list shown consist of the times,in seconds,that it took each of 10 school children to run a distance of 400 meter. If the SD of ten running times is 22.4 seconds, rounded to nearest tenth of second, how many of the 10 running times are more than one SD below the mean of the 10 running times ?

a. one b. two c. three d. four e. five

"How many of the 10 running times are more than one SD below the mean" means how many data points from given 10 are less than mean-1SD.

We are given that SD=22.4, so we should find mean --> mean=100 --> there are only 2 data points below 100-22.4=77.6, namely 70 and 75.

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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23 Sep 2010, 02:48

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Quick way to find the mean for the given #s....

If you a=observe carefully, the firt 5 numbers (70,75,80,85,90) are evenly spaced by distance 5..so mean for thease five #s is the middile # 80.

Last five #s are also spaced evenly but in an indirect manner.. 105 105 130 130 130 take 10 points from middle 130 and distribute 5 eac to the first 2 ==> 110 110 120 130 130 ==mean is the middle # 120 as the distancce from 110 to 120 is same as that from 120 to 130

Hence the toal mean is the average of 80 and 120 = 100

Hope it helps

Remember: GMAT never wants you to sum up the #s. It wants you to decode the given #s in the above or some other efficient way. Infact, it gives the set of #s which can easily be decoded.

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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23 Sep 2010, 10:21

ha what wordings, I got confused....read, re-read and than got answer B.

yesterday also I got a question which had very confusing wordings..... "which of these is MORE NOT LESS LIKELY to be" ----- what does that mean? sorry for deviating from the original question.... _________________

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Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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09 Oct 2010, 18:07

B

I was also looking for a quicker way to find the mean... mularimba's method is ok but it will probably take the same amount of time or more to do the intermediate calc. _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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17 Jul 2011, 23:16

sushma0805 wrote:

70 75 80 85 90 105 105 130 130 130

The list shown consist of the times,in seconds,that it took each of 10 school children to run a distance of 400 meter.If the SD of ten running times is 22.4 seconds,rounded to nearest tenth of second,how many of the 10 running times are more than one SD below the mean of the 10 running times ?

a. one b. two c. three d. four e. five

Is this the correct way to find mean... Arrange number ...take first and last number/2

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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17 Jul 2011, 23:46

siddhans wrote:

sushma0805 wrote:

70 75 80 85 90 105 105 130 130 130

The list shown consist of the times,in seconds,that it took each of 10 school children to run a distance of 400 meter.If the SD of ten running times is 22.4 seconds,rounded to nearest tenth of second,how many of the 10 running times are more than one SD below the mean of the 10 running times ?

a. one b. two c. three d. four e. five

Is this the correct way to find mean... Arrange number ...take first and last number/2

i.e 70 + 130 /2 = 200/2 = 100?

No. That should be done only with evenly spaced data set OR Arithmetic Progression.

2,4,6,8,10---Common Difference=2 constant 70 75 80 85 90 105 105 130 130 130-> Common difference is not constant. _________________

Re: 70 75 80 85 90 105 105 130 130 130 The list shown consist of [#permalink]

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18 Jul 2011, 00:01

fluke wrote:

siddhans wrote:

sushma0805 wrote:

70 75 80 85 90 105 105 130 130 130

The list shown consist of the times,in seconds,that it took each of 10 school children to run a distance of 400 meter.If the SD of ten running times is 22.4 seconds,rounded to nearest tenth of second,how many of the 10 running times are more than one SD below the mean of the 10 running times ?

a. one b. two c. three d. four e. five

Is this the correct way to find mean... Arrange number ...take first and last number/2

i.e 70 + 130 /2 = 200/2 = 100?

No. That should be done only with evenly spaced data set OR Arithmetic Progression.

2,4,6,8,10---Common Difference=2 constant 70 75 80 85 90 105 105 130 130 130-> Common difference is not constant.

Fluke,

Looks like i got a fluke here Just kidding ... So whats the fastest way of finding mean here?

Yes , this shudnt take u more than 25 secs. or u can assume mean = 130 use the principle of mean : sum of the deviations from the mean should always be zero

-60-55-50-45-40-25-25+0+0+0 = -300/10 = -30 so mean is 130-30 = 100

Here is how I computed the mean. It was pretty fast, and did not involve too many calculations: 70 75 80 85 90 105 105 130 130 130

First term (70 (100-30)) and last term (130 (100+30)) : Mean is 100. Second term (75 = 70 + 5= (100-30 + 5)) and 9th term (130): Mean is 100. 5 is carried over later. Third term (80 = 70 + 10 = (100-30 + 10)) and 8th term (130): Mean is 100. 10 is carried over later. Leave the fourth term (85) for the moment. Take the fifth term (90) and add (5+10) from earlier to make a total of 105. Then add this 105 with the remaining two 105 (6th and 7th terms), along with the fourth term to get: Mean = \(\frac{(105*3+85)}{4}\) = 100.

Though the explanation is kind of long, the computation is faster than computing by addition and multiplication of all terms. _________________

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