S1 help!

[Alla' Abu-Sultaneh]

I've been stuck on part c! help!
Jan 2005 Q7

 

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I've been stuck on part c! help!
Jan 2005 Q7

Given that P(X >= 79 + b) = 2P(X =< 79 – a),

(c) show that the area of the shaded region is 0.1179.

this one right???

 

[Alla' Abu-Sultaneh]

Given that P(X >= 79 + b) = 2P(X =< 79 – a),

(c) show that the area of the shaded region is 0.1179.

this one right???

Yes!

 

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Yes!

here you go... this might help you..

 

[Alla' Abu-Sultaneh]

here you go... this might help you..

Thank you soooooooooooooooo much <3

 

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Thank you soooooooooooooooo much <3

u'r welcome... tell me if any need any more help...

 

[saba1996]

THANK U SO MUCH! I need help in the same ppr Q 3 f) and g)

(f ) Find the values of (y bar) ± 1.96s.

(2)​

(g) Comment on whether or not you think that the weights of these students could be modelled by a normal distribution.

(1)​

 

[Alla' Abu-Sultaneh]

THANK U SO MUCH! I need help in the same ppr Q 3 f) and g)

(f ) Find the values of (y bar) ± 1.96s.

(2)​

(g) Comment on whether or not you think that the weights of these students could be modelled by a normal distribution.

(1)​

Sorry for the late reply! for find the mean of Y , and then add to it 1.96 multiplied by the standard deviation, then do the same thing but instead of adding, subtract 1.96*standard deviation

 

[saba1996]

Sorry for the late reply! for find the mean of Y , and then add to it 1.96 multiplied by the standard deviation, then do the same thing but instead of adding, subtract 1.96*standard deviation

THANKS A BUNCHHH

 

[saba1996]

here you go... this might help you..

Hey ! i need help in S1!!!!! Jan 2009 Q2 b) and c) and Q3 e) PLEASEEEEEEEE> http://www.edexcel.com/migrationdoc...um 2000/January 2009/6683_01_que_20090119.pdf

 

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Hey ! i need help in S1!!!!! Jan 2009 Q2 b) and c) and Q3 e) PLEASEEEEEEEE> http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/January 2009/6683_01_que_20090119.pdf

for b part: find {{ 1 - P(B ∪ E) }} by [[ 1 - { P(B ∩ E') + P(B ∩ E) + P(B' ∩ E) } ]]

= [[ 1 - { (2/5) - x } + { x } + { (2/3) - x }​

= 1 - {(2/5) - x + x + (2/3) - x } <<Underlines expressions are cancelled out>>​

= 1 - (2/5) - (2/3) + x​

= 1 - (2/5) - (2/3) + (6/25) << x = (6/25)>> from part (a)​

= 0.173 (to 3 sig. fig.)​

​

for c part: to show statistically independent by using [[ P(E | B) = P(E) ]]

P(E | B) = (9/25) = 0.36​

P(E) = (2/5) = 0.40​

P(E | B) is not equal to P(E)​

So E and B are not statistically independent.​

 

[saba1996]

for b part: find {{ 1 - P(B ∪ E) }} by [[ 1 - { P(B ∩ E') + P(B ∩ E) + P(B' ∩ E) } ]]

= [[ 1 - { (2/5) - x } + { x } + { (2/3) - x }​

= 1 - {(2/5) - x + x + (2/3) - x } <<Underlines expressions are cancelled out>>​

= 1 - (2/5) - (2/3) + x​

= 1 - (2/5) - (2/3) + (6/25) << x = (6/25)>> from part (a)​

= 0.173 (to 3 sig. fig.)​

​

for c part: to show statistically independent by using [[ P(E | B) = P(E) ]]

P(E | B) = (9/25) = 0.36​

P(E) = (2/5) = 0.40​

P(E | B) is not equal to P(E)​

So E and B are not statistically independent.​

THANK YOU SOOOO MUCH!

 

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THANK YOU SOOOO MUCH!

ur welcum