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The list shown consist of the times, in seconds, that it [#permalink]

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28 Mar 2010, 09:42

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70, 75, 80, 85, 90, 105, 105, 130, 130, 130 The list shown consists of the times, in seconds, that it took each of 10 schoolchildren to run a distance of 400 on of meters. If the standard devastation of the 10 running times is 22.4 seconds, rounded to the nearest tenth of a second, how many of the 10 running times are more than 1 standard deviation below the mean of the 10 running times?

the mean of the 10 running times is 100 sec. we need to find how many of the recorded times are less than 1 SD below the mean i.e how many took less than 100 sec - 1SD = 100 sec - 22.4 sec = 77.6sec. The times are 70 and 75. hence 2

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: The list shown consist of the times, in seconds, that it [#permalink]

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03 Jun 2013, 15:23

To find the mean, take a wild stab and pick a number within the range provided by the given numbers. Now, find the delta between that number and each number in the set. So now you get the overall delta from your "assumed mean". Split that up equally among each number. So adding that to your "assumed mean" will give you the actual mean.

I took 100 and found the difference between it and each of the numbers in the set. Turns out that the overall delta neatly becomes 0. So 100 is the actual mean.

Re: The list shown consist of the times, in seconds, that it [#permalink]

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04 Jun 2013, 03:39

2

This post received KUDOS

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

The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake. it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90; Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-22,4=77,6. There are only two numbers below 77,6. The answer is B
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If you found my post useful and/or interesting - you are welcome to give kudos!

Re: The list shown consist of the times, in seconds, that it [#permalink]

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03 Sep 2013, 23:55

ziko 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

The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake. it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90; Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-22,4=77,6. There are only two numbers below 77,6. The answer is B

Pardon me for my understanding, but "are <more than> <one SD below the mean> of the 10 running times? So isn't it asking values greater than 77.6 ?

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

The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake. it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90; Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-22,4=77,6. There are only two numbers below 77,6. The answer is B

Pardon me for my understanding, but "are <more than> <one SD below the mean> of the 10 running times? So isn't it asking values greater than 77.6 ?

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: The list shown consist of the times, in seconds, that it [#permalink]

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04 Sep 2013, 02:13

Bunuel wrote:

ygdrasil24 wrote:

ziko 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

The most time consuming part in this question is to define the mean. Under exam pressure and time pressure it is very easy to make mistake. it is easier to group numbers: 130*3=390; 105*2=210; 75+85=160; 70+80=150; 90; Next stage combine results, again using more convenient ways to calculate: 390+210=600; 160+150=310; 90. 600+310+90=1000. Since there are 10 numbers the mean is 100. Questions asks to find the quantity of numbers one SD BELOW the mean, which is 100-22,4=77,6. There are only two numbers below 77,6. The answer is B

Pardon me for my understanding, but "are <more than> <one SD below the mean> of the 10 running times? So isn't it asking values greater than 77.6 ?

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.

Why couldnt they have written it as How many of the 10 running times are less than one SD below the mean. This is frustrating for me

Re: The list shown consist of the times, in seconds, that it [#permalink]

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14 Oct 2014, 06:01

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Re: The list shown consist of the times, in seconds, that it [#permalink]

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13 Jul 2016, 16:59

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: The list shown consist of the times, in seconds, that it [#permalink]

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13 Jul 2016, 17:24

Instead of wasting valuable time adding up all 10 numbers, you can spot some easy clues the GMAT leaves behind. The first 5 numbers are perfectly symmetric around 80, so you know mean of the first 5 is 80. The second half of numbers can be quickly added together- 3*130 = 390 and 105*2 = 210 for a sum of 600. Divide by 5 to get 120. Since the first half has a mean of 80 and second half has a mean of 120, and both halves have equal weight (they both represent 5 integers), the average of both halves will be right in the middle at 100. From there figuring out the answer is business as usual.

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