# maths a level help- statistics-binomial distribution

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Thread starter 4 weeks ago
#1

just wanted to know if how im doing it is correct,
so for the first part i used simutanous equation:
i found the propablity of the z values which are -1.9597 and 0.6745
so its -1.9507 = (15-u)/s.d and 0.6745 = (20-u)/s.d
i get s.d as 1.898 and u as 18.7, is this the right way of going about it, i think another way you could do it is using the graph but im not sure how? could someone explain this way to me

for the second part:
y = 2x+4, so for the mean you x 2 and then +4
and for the s.d? apparently you x 4? becuase you do 2^2? wouldnt you just times it by 2?
then it becomes X~B (41.4,square root(7.6)^2)
Last edited by r.maliak; 4 weeks ago
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4 weeks ago
#2
For the first part, you can write the mean straight down because of the symmetry in the given info If not, solve the simultaneous equations for mu and sigma.
Then get the standard deviation, sigma, from the z value. Dont know what the 0.6 represents, but agree with the -1.9. That means the point lies ~1.9 standard deviations from the mean so its easy to get sigma.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#3
(Original post by r.maliak)

just wanted to know if how im doing it is correct,
so for the first part i used simutanous equation:
i found the propablity of the z values which are -1.9597 and 0.6745
so its -1.9507 = (15-u)/s.d and 0.6745 = (20-u)/s.d
i get s.d as 1.898 and u as 18.7, is this the right way of going about it, i think another way you could do it is using the graph but im not sure how? could someone explain this way to me

for the second part:
y = 2x+4, so for the mean you x 2 and then +4
and for the s.d? apparently you x 4? becuase you do 2^2? wouldnt you just times it by 2?
then it becomes X~B (41.4,square root(7.6)^2)
If you look at the way variance is defined and what it means, the you would notice:

* it measures how spread out data is … so shifting the data by +4 units wont affect how spread out it is therefore +4 has no effect on variance

* variance is related to SQUARED data … so if you multiply data by 2 the effect is that your variance gets multiplied by 2^2.

Generally; Var(aX+b) = a^2 Var(X)
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4 weeks ago
#4
(Original post by r.maliak)

just wanted to know if how im doing it is correct,
so for the first part i used simutanous equation:
i found the propablity of the z values which are -1.9597 and 0.6745
so its -1.9507 = (15-u)/s.d and 0.6745 = (20-u)/s.d
i get s.d as 1.898 and u as 18.7, is this the right way of going about it, i think another way you could do it is using the graph but im not sure how? could someone explain this way to me

for the second part:
y = 2x+4, so for the mean you x 2 and then +4
and for the s.d? apparently you x 4? becuase you do 2^2? wouldnt you just times it by 2?
then it becomes X~B (41.4,square root(7.6)^2)
Also why is Y=2X+4 distributed binomially?
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Thread starter 4 weeks ago
#5
(Original post by mqb2766)
For the first part, you can write the mean straight down because of the symmetry in the given info If not, solve the simultaneous equations for mu and sigma.
Then get the standard deviation, sigma, from the z value. Dont know what the 0.6 represents, but agree with the -1.9. That means the point lies ~1.9 standard deviations from the mean so its easy to get sigma.
with the 0.6, i thaught the z value always has to be on the left hand side, but with x> 20 its on the right?
so instead i did 1-0.025 to get the left hand side and put this on the calc to get P

17.5? right, I must have made an error with the simultaneous equation becuase i ddnt get the same answer
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Thread starter 4 weeks ago
#6
(Original post by RDKGames)
Also why is Y=2X+4 distributed binomially?
im not sure but thx for the help from you other post
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4 weeks ago
#7
(Original post by r.maliak)
with the 0.6, i thaught the z value always has to be on the left hand side, but with x> 20 its on the right?
so instead i did 1-0.025 to get the left hand side and put this on the calc to get P

17.5? right, I must have made an error with the simultaneous equation becuase i ddnt get the same answer
17.5 is the mean. And you can use the Z=-1.9 point, X=15 to get the sigma, but the Z score is ~1.9 for the X=20 point. There is an obvious symmetry with the -1.9 for the X=15 point. Both are approximately two standard deviations from the mean. So the standard deviation is about 2.5/2

I still don't understand how you get Z=0.6 for X=20. It is to the right of the mean as Z is positive.
Last edited by mqb2766; 4 weeks ago
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Thread starter 4 weeks ago
#8
(Original post by RDKGames)
If you look at the way variance is defined and what it means, the you would notice:

* it measures how spread out data is … so shifting the data by +4 units wont affect how spread out it is therefore +4 has no effect on variance

* variance is related to SQUARED data … so if you multiply data by 2 the effect is that your variance gets multiplied by 2^2.

Generally; Var(aX+b) = a^2 Var(X)
oh wait what im confised about is do you write it in the form X~N (u, sqaure root(s.d)^2
so am i not finding the s.d and not variance?
my teacher found the variance and wrote is as (u, sqaure root(variance)^2
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4 weeks ago
#9
(Original post by r.maliak)
oh wait what im confised about is do you write it in the form X~N (u, sqaure root(s.d)^2
so am i not finding the s.d and not variance?
my teacher found the variance and wrote is as (u, sqaure root(variance)^2
the standard deviation is ~1.25 so the variance, which the question asks for, is ~1.25^2. Im not clear about what you/your teacher wrote. You normally use the standard deviation with the z value (sqrt(variance)), but quote the variance (stddev^2) when you define the normal distribution.

It would help if you upload your (careful) working and its easier to give proper feedback.
Last edited by mqb2766; 4 weeks ago
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4 weeks ago
#10
you write the distribution as

X ~ N ( µ, σ2 )

or

X ~ N ( mean, variance )
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Thread starter 4 weeks ago
#11
(Original post by mqb2766)
the standard deviation is ~1.25 so the variance is ~1.25^2. Im not clear about what you/your teacher wrote.
It would help if you upload your (careful) working and its easier to give proper feedback.
so the mean is 17.5 ,if you put this back into the equation you get -1.9597 = 15-17.5/s.d
s.d = 1.27
v = 1.27^2 = 1.6
so would you write it like X~N( 17.5 , sqaure root 1.6^2)
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4 weeks ago
#12
(Original post by r.maliak)
so the mean is 17.5 ,if you put this back into the equation you get -1.9597 = 15-17.5/s.d
s.d = 1.27
v = 1.27^2 = 1.6
so would you write it like X~N( 17.5 , sqaure root 1.6^2)
N(17.5, 1.6)
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Thread starter 4 weeks ago
#13
(Original post by mqb2766)
N(17.5, 1.6)
oh the sqaure root and sqaure cancel out anways ok i get it now thx
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4 weeks ago
#14
(Original post by r.maliak)
oh the sqaure root and sqaure cancel out anways ok i get it now thx
Yes. Note the comment in the previous post about using the standard deviation when using the Z score, thinking about the width of the distrubtion, ... etc, but quoting the variance when defining the normal distribution. Its a common thing to get confused about, but really there is little to it, just square or square root to swap between the two.
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Thread starter 4 weeks ago
#15
(Original post by mqb2766)
17.5 is the mean. And you can use the Z=-1.9 point, X=15 to get the sigma, but the Z score is ~1.9 for the X=20 point. There is an obvious symmetry with the -1.9 for the X=15 point. Both are approximately two standard deviations from the mean. So the standard deviation is about 2.5/2

I still don't understand how you get Z=0.6 for X=20. It is to the right of the mean as Z is po
oh i checked the numbers on the calc again and i didnt put 0.975 but 0.75 my bad, it gives 1.95 Last edited by r.maliak; 4 weeks ago
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4 weeks ago
#16
(Original post by r.maliak)
oh i checked the numbers on the calc again and i didnt put 0.975 but 0.75 my bad, but it gives 1.95 Good its sorted. Advice is to sketch the distribution and understand what the Z value represents. For this problem it should be clear that -1.95 was correct (2.5% area in the tail) and the 0.6 was incorrect, even just by symmetry. Otherwise youre just plugging values into a calc and when the inevitable mistake happens, you don't have a way of spotting it.
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Thread starter 4 weeks ago
#17
(Original post by mqb2766)
Good its sorted. Advice is to sketch the distribution and understand what the Z value represents. For this problem it should be clear that -1.95 was correct (2.5% area in the tail) and the 0.6 was incorrect, even just by symmetry. Otherwise youre just plugging values into a calc and when the inevitable mistake happens, you don't have a way of spotting it.
aaah ok i get what z values mean now and symmetry, I appreciate the time taken to help me!
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