# maths a level help- statistics-binomial distribution

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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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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.

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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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)

**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)

* 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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**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)

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(Original post by

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.

**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.

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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(Original post by

Also why is Y=2X+4 distributed binomially?

**RDKGames**)Also why is Y=2X+4 distributed binomially?

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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

**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

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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(Original post by

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)

**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)

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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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

**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

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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(Original post by

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.

**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.

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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#12

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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)

**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)

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(Original post by

N(17.5, 1.6)

**mqb2766**)N(17.5, 1.6)

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#14

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oh the sqaure root and sqaure cancel out anways ok i get it now thx

**r.maliak**)oh the sqaure root and sqaure cancel out anways ok i get it now thx

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(Original post by

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

**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

Last edited by r.maliak; 4 weeks ago

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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

**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

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(Original post by

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.

**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.

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