# Range of one standard deviation from mean

1. Jun 14, 2014

### kelvin macks

my question is on part ii, my range ( as shown in the photo), is between 4.52 and 16.28, so my working would be (6+10+4+4)/(6+10+4+4+1) X100% ... the answer given is 72% which is differnt form my answer? which part of my working is wrong ?

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2. Jun 14, 2014

### Simon Bridge

Data: # PCs sold in 1 week over 25 weeks:
10,8,7,4,4,8,9,14,17,11,5,4,5,7,10,12,14,10,9,29,7,6,5,10,15

You are asked to find the mean and standard deviation.
you have 1 sd range of 4.52-16.28 which suggests you got a mean of 10.4 and a std dev of 4.8.
Does this agree with the model answers?

Then you are asked to find the percentage of weeks where the number sold per week is within 1sd from the mean.

Your working seems inconsistent with this.
i.e. the way I read it:
If there were 5 weeks where the number sold was 1sd from mean, then the calculation would go 100x5/25=20%
Because there are 25 weeks in total and 5 of them fit inside the range.

3. Jun 15, 2014

### kelvin macks

my answer for standard deviation and mean are correct.. do you mean i should get the ans from the data 10,8,7,4,4,8,9,14,17,11,5,4,5,7,10,12,14,10,9,29,7,6,5,10,15 ? and not from the grouped data (1-5), (6-10) and etc ? my ans now is 18/25 now!

4. Jun 15, 2014

### Simon Bridge

OK - so the range is: 5.6 to 15.2

Using data
10,8,7,4,4,8,9,14,17,11,5,4,5,7,10,12,14,10,9,29,7,6,5,10,15

Numbers 6-15 inclusive are in the range. Those are:
10,8,7,8,9,14,11,7,10,12,14,10,9,7,6,10,15
17/25 = 68%

Which is fair - you should be inside 1sd of the mean about 68% of the time for normally distributed data.

18/25 is 72%, which agrees with the model answer - did I accidentally exclude one?