How do I tackle standard deviation questions UNIT 2 AQA AS Watch

TheJCEP
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Really struggling on standard deviation questions.

What I understand that standard deviation is the spread around the mean.

Would someone tell me about me standard deviation and the best way actually tackling a standard deviation questions and points to mention would really appreciate it.
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the bear
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http://www.robertniles.com/stats/stdev.shtml

https://www.youtube.com/watch?v=hUaua15QzK4
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steph996
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I was struggling with this myself massively until a few days ago, so i thought id share with you how i learnt it
say for example we have 3 different antibiotics tested on a petri dish of bacteria , and the value given is the amount of clear zone around the disks soaked in the different antibiotics...


MEAN CLEAR ZONE IN MM/ SD:
A - 22.8 ± 3.9
B -9.1 ± 2.6
C- 6.6 ± 1.5

you and and minus the number on the right (which is the standard deviation to find the RA) to your original value
so A's range = 18.9 to 26.7
Bs range would be = 6.5 to 11.7
C's range would be = 5.1 to 8.1

if any of the numbers overlap with each other, that means the error bars on a graph would overlap with each other and there is NO significant difference between these two values/ however many values overlap

this means there is no significant difference between B& C as their ranges meet

as you can see A doesn't overlap with B&C which overlap with each other meaning that there is a significant difference between A and the rest of the values and it is unlikely that the results are due to chance

meaning you would say that A is the most affective antibiotic because it has the larges space cleared and there is a significant difference between A compared to B&C...

hope this helps!
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TheJCEP
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(Original post by steph996)
I was struggling with this myself massively until a few days ago, so i thought id share with you how i learnt it
say for example we have 3 different antibiotics tested on a petri dish of bacteria , and the value given is the amount of clear zone around the disks soaked in the different antibiotics...


MEAN CLEAR ZONE IN MM/ SD:
A - 22.8 ± 3.9
B -9.1 ± 2.6
C- 6.6 ± 1.5

you and and minus the number on the right (which is the standard deviation to find the RA) to your original value
so A's range = 18.9 to 26.7
Bs range would be = 6.5 to 11.7
C's range would be = 5.1 to 8.1

if any of the numbers overlap with each other, that means the error bars on a graph would overlap with each other and there is NO significant difference between these two values/ however many values overlap

this means there is no significant difference between B& C as their ranges meet

as you can see A doesn't overlap with B&C which overlap with each other meaning that there is a significant difference between A and the rest of the values and it is unlikely that the results are due to chance

meaning you would say that A is the most affective antibiotic because it has the larges space cleared and there is a significant difference between A compared to B&C...

hope this helps!
Thank you, thank you thank you thank you thank you!!!!!!!!!!!
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Bhavika22
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(Original post by steph996)
i was struggling with this myself massively until a few days ago, so i thought id share with you how i learnt it
say for example we have 3 different antibiotics tested on a petri dish of bacteria , and the value given is the amount of clear zone around the disks soaked in the different antibiotics...


Mean clear zone in mm/ sd:
A - 22.8 ± 3.9
b -9.1 ± 2.6
c- 6.6 ± 1.5

you and and minus the number on the right (which is the standard deviation to find the ra) to your original value
so a's range = 18.9 to 26.7
bs range would be = 6.5 to 11.7
c's range would be = 5.1 to 8.1

if any of the numbers overlap with each other, that means the error bars on a graph would overlap with each other and there is no significant difference between these two values/ however many values overlap

this means there is no significant difference between b& c as their ranges meet

as you can see a doesn't overlap with b&c which overlap with each other meaning that there is a significant difference between a and the rest of the values and it is unlikely that the results are due to chance

meaning you would say that a is the most affective antibiotic because it has the larges space cleared and there is a significant difference between a compared to b&c...

Hope this helps!
thank you so much 😊
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jimmarshall
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Ques: Consider the following three data sets A, B and C.

A = {9,10,11,7,13}

B = {10,10,10,10,10}Find

C = {1,1,10,19,19}

a) Calculate the mean of each data set.

b) Calculate the standard deviation of each data set.

c) Which set has the largest standard deviation?

d) Is it possible to answer question c) without calculations of the standard deviation?

Solution -
(a) mean of Data set A = (9+10+11+7+13)/5 = 10
mean of Data set B = (10+10+10+10+10)/5 = 10
mean of Data set C = (1+1+10+19+19)/5 = 10

(b) Standard Deviation Data set A= √[ ( (9-10)2+(10-10)2+(11-10)2+(7-10)2+(13-10)2 )/5 ] = 2
Standard Deviation Data set B= √[ ( (10-10)2+(10-10)2+(10-10)2+(10-10)2+(10-10)2 )/5 ] = 0
Standard Deviation Data set C= √[ ( (1-10)2+(1-10)2+(10-10)2+(19-10)2+(19-10)2 )/5 ] = 8.05

(c) Data set C has the largest standard deviation.

(d) Yes, since data Set C has data values that are further away from the mean compared to sets A and B.


Ques -
A given data set has a mean μ and a standard deviation σ.

a) What are the new values of the mean and the standard deviation if the same constant k is added to each data value in the given set?Explain.

b) What are the new values of the mean and the standard deviation if each data value of the set is multiplied by the same constant k?Explain.

Solution -

(a)
We limit the discussion to a data set with 3 values for simplicity, but the conclusions are true for any data set with quantitative data.

Let x, y and z be the data values making a data set.
The mean μ = (x + y + z) / 3
The standard deviation σ = √[ ((x - μ)2 + (y - μ)2 + (z - μ)2)/3 ]
We now add a constant k to each data value and calculate the new mean
μ'. μ' = ((x + k) + (y + k) + (z + k)) / 3 = (x + y + z) / 3 + 3k/3 = μ + k
We now calculate the new mean standard deviation σ'.
σ' = √[ ((x + k - μ')2 +(y + k - μ')2+(z + k - μ')2)/3 ]
Note that x + k - μ' = x + k - μ - k = x - μ
also y + k - μ' = y + k - μ - k = y - μ and z + k - μ' = z + k - μ - k = z - μ
Therefore σ' = √[ ((x - μ)2 +(y - μ)2+(z - μ)2)/3 ] = σ
If we add the same constant k to all data values included in a data set, we obtain a new data set whose mean is the mean of the original data set PLUS k. The standard deviation does not change.

(b)
We now multiply all data values by a constant k and calculate the new mean μ' and the new standard deviation σ'.
μ' = (kx + ky + kz) / 3 = kμ
σ' = √[ ((kx - kμ)2 +(ky - kμ)2+(kz - kμ)2)/3 ] = |k| σ
If we multiply all data values included in a data set by a constant k, we obtain a new data set whose mean is the mean of the original data set TIMES k and standard deviation is the standard deviation of the original data set TIMES the absolute value of k.
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hum1
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http://us19.chatzy.com/71814616680493

biology chat group to fire questiions at each other for the exam...
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steph996
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You're welcome guys, best of luck in your exams!
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