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A vending machine is designed to dispense 8 ouces of coffee

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A vending machine is designed to dispense 8 ouces of coffee [#permalink] New post 27 Nov 2006, 18:13
A vending machine is designed to dispense 8 ouces of coffee into a cup. After a test that recorded the number of ounces in a cup of coffee in each of 1,000 cups dispensed by the vending machine, the 12 listed amounts in ounces, (see below) were selected from the data. If the 1,000 recorded amounts have a mean of 8.1 and a standard deviation of .3 ounce, how many of the 12 listed amounts are within 1.5 standard deviations of the mean?

7.51 8.22 7.86 8.3
8.09 7.83 7.96 8.36
7.73 8.25 8.01 8.53

A) 4
B) 6
C) 9
D) 10
E) 11

Argh - I can't figure out what I am doing wrong...Shouldn't the standard deviation yied a range of 7.8 - 8.4 hence 2 amounts are outside the standard deviation?? Please help
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 [#permalink] New post 27 Nov 2006, 19:02
E for me

Mean = 8.1
SD = 0.3

Range of values within 1.5 SDs of mean is given by 7.65-8.55

11 values fall within this range
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 [#permalink] New post 28 Nov 2006, 00:43
Yes E....

Simply multiply 1.5*0.3=0.45

So the lowest possible value can be: 8.1-0.45=7.65
The highers possible value: 8.1+0.45=8.55
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 [#permalink] New post 28 Nov 2006, 16:27
SimaQ wrote:
Yes E....

Simply multiply 1.5*0.3=0.45

So the lowest possible value can be: 8.1-0.45=7.65
The highers possible value: 8.1+0.45=8.55


Thanks SimaQ - That helps! E it is
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Re: [#permalink] New post 04 Jan 2009, 19:31
Why do we need to multiple 1.5 * .3?



SimaQ wrote:
Yes E....

Simply multiply 1.5*0.3=0.45

So the lowest possible value can be: 8.1-0.45=7.65
The highers possible value: 8.1+0.45=8.55
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Re: Re: [#permalink] New post 04 Jan 2009, 20:37
bigfernhead wrote:
Why do we need to multiple 1.5 * .3?



SimaQ wrote:
Yes E....

Simply multiply 1.5*0.3=0.45

So the lowest possible value can be: 8.1-0.45=7.65
The highers possible value: 8.1+0.45=8.55


because you are being asked for "within 1.5 standard deviations of the mean". Standard deviation is 0.3, 1.5 standard deviations is 0.45.
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Re: GMAT Prep - Standard Deviation [#permalink] New post 05 Jan 2009, 05:09
The range is always
M + factor * SD
M - factor * SD
Re: GMAT Prep - Standard Deviation   [#permalink] 05 Jan 2009, 05:09
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