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# Stats question watch

1. i am really stuck on the following question can some one help me out:

here are 29 measurements given below:

4.07 4.88 5.10 5.26 5.27 5.29 5.29 5.30 5.34 5.34 5.36 5.39 5.42 5.44 5.46 5.47 5.50 5.53 5.55 5.57 5.58 5.61 5.62 5.63 5.65 5.75 5.79 5.85 5.86

if the data were standardised to a scale of mean zero and standard devation 1:
i)what would the smallest standardised value be?
ii) what value would correspond to a standardised value of +2.6?
2. (Original post by az1)
i am really stuck on the following question can some one help me out:

here are 29 measurements given below:

4.07 4.88 5.10 5.26 5.27 5.29 5.29 5.30 5.34 5.34 5.36 5.39 5.42 5.44 5.46 5.47 5.50 5.53 5.55 5.57 5.58 5.61 5.62 5.63 5.65 5.75 5.79 5.85 5.86

if the data were standardised to a scale of mean zero and standard devation 1:
i)what would the smallest standardised value be?
ii) what value would correspond to a standardised value of +2.6?
zero?
2.6?
3. yep zero and 2.6. that's wot it says on the practise exam paper and like you i too am confused.
4. (Original post by az1)
i am really stuck on the following question can some one help me out:

here are 29 measurements given below:

4.07 4.88 5.10 5.26 5.27 5.29 5.29 5.30 5.34 5.34 5.36 5.39 5.42 5.44 5.46 5.47 5.50 5.53 5.55 5.57 5.58 5.61 5.62 5.63 5.65 5.75 5.79 5.85 5.86

if the data were standardised to a scale of mean zero and standard devation 1:
i)what would the smallest standardised value be?
ii) what value would correspond to a standardised value of +2.6?
What the question wants you to do is use the normal distribution. first we find the mean (lets call it m) and the standard deviation (lets call it s) of the measurements. then for each measurement, we minus m and divide by s to get the standardised value. The smallest value will be (4.07-m)/s

ii). let unknown value be x. (x-m)/s=+2.6 and you work out what x is from there.

hope this helps!
5. To standardise a value, you take away the mean then divide by the standard deviation.

Xstandardised=(Xraw-mean)/σ

The mean is 5.42 and the stadard deviation is 0.333 (to 3sf).

As the lowest value before standardisation is 4.07, the lowest value after standardisation will be this one when standardised:

(4.07-5.42)/0.333=-3.75 (to 3sf)

For the second part, you can rearrange the formula to get:
Xstandardised+mean=Xraw

So the value is: 2.6x0.333+5.42=6.29 (to 3sf)
6. (Original post by az1)
yep zero and 2.6. that's wot it says on the practise exam paper and like you i too am confused.
omg that was such an instant guess and i got it right
7. (Original post by meepmeep)
To standardise a value, you take away the mean then divide by the standard deviation.

Xstandardised=(Xraw-mean)/σ

The mean is 5.42 and the stadard deviation is 0.333 (to 3sf).

As the lowest value before standardisation is 4.07, the lowest value after standardisation will be this one when standardised:

(4.07-5.42)/0.333=-3.75 (to 3sf)

For the second part, you can rearrange the formula to get:
Xstandardised+mean=Xraw

So the value is: 2.6x0.333+5.42=6.29 (to 3sf)
mean of zero...
8. btw its always better to look at the no. of marks b4 doing a question... ppl always think of stuff more than needed.
9. (Original post by z!D4N)
mean of zero...
The mean of the data is the sum divided by the number of pieces of data, so it can't be zero in this case, as all of the pieces of data are greater than 0.

When standardised, it becomes 0, but that's only because, as p8224 said, (x-m)/s gives the stadardised value and so it's the mean minus itself.

Good advice on exam technique though. Don't got bogged down in a question if it's not worth many marks and you're going nowhere slowly.
10. thanks for the help and btw it was worth 4 marks
11. (Original post by az1)
thanks for the help and btw it was worth 4 marks
that seems about right. one mark for finding the mean, one for s.d...and one each for parts i) and ii). i dont think it's too much work to do for 4 marks...
but like others have said..if you're stuck, just move on.

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