As level s1 statistics question
Can someone please help me with this question, the attached image is below
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Original post by Universecolors
Can someone please help me with this question, the attached image is below
Not this one , but the one below it, I made a mistakes , my question is in the picture below this one
(edited 5 years ago)
Sorry the one , 7 ii
Not clear which question , remove the other images to the one you need 
Original post by MatthewAV8B
Not clear which question , remove the other images to the one you need
This one please 7 ii
(edited 5 years ago)
Okay so :
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
Original post by MatthewAV8B
Okay so :
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
You explained this so nicely , I can’t thank you enough , can you please tell me why for part b , the same question , the 0.8 is negative
Original post by Universecolors
You explained this so nicely , I can’t thank you enough , can you please tell me why for part b , the same question , the 0.8 is negative
Sure thing , just bear with me as I do some scribbles
Original post by MatthewAV8B
Okay so :
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
We know that P ( 35.2 < X < t ) = 0.148
We can split this up into two separate probabilities by looking at a normal distribution curve and seeing what we are after.
P ( x < t ) - P( 35.2 < x ) = 0.148
Now for the second bracket [ P( 35.2 < x ) ] , we can find a given probability using a formula
z = Our Value - Mean / Standard Deviation
This gives us : z = 35.2-35.2 / 4.7
Which makes z = 0
Following from this value , we take the given Z value and go to our tables and look up the given value for when z = 0 :
Which nicely makes 0.5!!!
From there , we can re insert this value into our equation.
P( x < t ) - 0.5 = 0.148
Thus : P( x < t ) = 0.648 [ As the mark scheme wants ]
With this probability , we go back to the tables and find the value for Z we need to calculate T
This comes out as 0.380
Thus using z = (X - Mean) / Standard deviation
0.380 = (t - 35.2) / 4.7
t = (0.380 x 4.7) + 35.2
Which then makes T = 37 when rounded , hope this helps
Thanks , ur helping me a lot , if u do (b) please tell me
Original post by Universecolors
You explained this so nicely , I can’t thank you enough , can you please tell me why for part b , the same question , the 0.8 is negative
Okay for part b)
We are told N( μ , σ^2)
This just means a normal distribution with Mean μ and Standard Deviation σ^2. They are checking that you understand the different notation.
We are also told the following :
P ( x < 7 ) = 0.2119 ------ Lets call this A
P ( x < 10) = 0.6700 -------- Lets call this B
From these , we go to our tables to get our given Z values :
For A
, we have a problem!!
, values < 0.5 are not tabulated! , but we can get around this using a simple trick!We know the normal distribution is symmetrical , so we can do:
1 - 0.2119 , giving us 0.7881
This is much better! , lets go to the tables :
Using 0.78 , we get the Z value of : 0.7722
Now remember, because we want this value , but at the other side of the distribution , we make this negative!
Giving the Z values for A as -0.7722 , we then round this to make our lives easier in later calculations
Thus Z value for A = -0.8 [As you asked!!]
For B
Because this value is > 0.5 , its tabulated and our lives are easier , we just plug and play with 0.67 into the tables :
Attachment not found
Giving us a Z for B as 0.4399 , which we round to 0.44
Now with our values , lets form some equations
Using Z = ( Value - mean ) / Standard Deviation
we make these two :
-0.8 = (7 - μ)/σ and 0.44 = (10 - μ)/σ
Expand the bottoms out
-0.8σ = 7 - μ and 0.44σ = 10 - μ
Lets put ontop of one another to see what we get
-0.8σ = 7 - μ
0.44σ = 10 - μ
Awesome! , We have simultaneous equation!
So now we can solve and get the values stated in the mark scheme :
μ = 8.94 and σ = 2.42
Hope this helps!
Original post by Universecolors
Thanks , ur helping me a lot , if u do (b) please tell me
Part B done , have a read
Original post by MatthewAV8B
Part B done , have a read
Alright I get it , thanks for explaining it that way but there’s something confusing me , so u are saying if the area is less than 0.5 then we have to do what u talked about and it becomes negative , but why doesn’t become negative for this question below , question 1
Original post by Universecolors
Alright I get it , thanks for explaining it that way but there’s something confusing me , so u are saying if the area is less than 0.5 then we have to do what u talked about and it becomes negative , but why doesn’t become negative for this question below , question 1
Can u please answ
Original post by MatthewAV8B
Part B done , have a read
Can u please answer my question : (
Original post by Universecolors
Can u please answ
Can u please answer my question : (
Can u please answer my question : (
Hey dude sorry for that , been studying today and missed the notification , give me couple of mins to get this
Original post by MatthewAV8B
Hey dude sorry for that , been studying today and missed the notification , give me couple of mins to get this
Ok thanksssss !
Original post by Universecolors
Ok thanksssss !
Okay my dude :
We are told the usual stuff , mean of 20 and Variance of 49
And importantly that P(X>7) = 0.25
The greater than is why its not negative.
We cant find a given Z score from a probability of 0.25 , so we use 1-0.25 , which is 0.75 , which we then look up in the percentage points table :
This gives us the value 0.674
then we use the usual , Z = (Our value - mean) / Standard deviation
thus 0.674 = (k - 20) / 7
( It has become 7 because we square rooted the Variance to get the Standard Deviation )
Which when solved , gives up k = 24.7
Hope this helps , sorry for slow reply homie
Original post by MatthewAV8B
Okay my dude :
We are told the usual stuff , mean of 20 and Variance of 49
And importantly that P(X>7) = 0.25
The greater than is why its not negative.
We cant find a given Z score from a probability of 0.25 , so we use 1-0.25 , which is 0.75 , which we then look up in the percentage points table :
This gives us the value 0.674
then we use the usual , Z = (Our value - mean) / Standard deviation
thus 0.674 = (k - 20) / 7
( It has become 7 because we square rooted the Variance to get the Standard Deviation )
Which when solved , gives up k = 24.7
Hope this helps , sorry for slow reply homie
We are told the usual stuff , mean of 20 and Variance of 49
And importantly that P(X>7) = 0.25
The greater than is why its not negative.
We cant find a given Z score from a probability of 0.25 , so we use 1-0.25 , which is 0.75 , which we then look up in the percentage points table :
This gives us the value 0.674
then we use the usual , Z = (Our value - mean) / Standard deviation
thus 0.674 = (k - 20) / 7
( It has become 7 because we square rooted the Variance to get the Standard Deviation )
Which when solved , gives up k = 24.7
Hope this helps , sorry for slow reply homie
No , I am the one to apologize for asking a lot of questions , so when it says greater that means it’s positivve ? What if it was less than , will that make it negative ? For example 7 i
Attachment not found
Here is answer
Attachment not found
(edited 5 years ago)
Original post by Universecolors
No , I am the one to apologize for asking a lot of questions , so when it says greater that means it’s positivve ? What if it was less than , will that make it negative ? For example 7 i
Here is answer
Attachment not found
Here is answer
Attachment not found
Questions are good , its how we learn.
Yeah because its greater than , we leave the value as the Raw 1 - whatever , because its not tabulated if its less than . Its hard to explain as I only ever did stats 1 , so I dont know all the detail like a teacher would.
If it was less than , we would need to do one minus our already one misused value if you understand , sorry this bit isnt as clear as the rest :P , ill look it up in the book for further info if you need.
All the best.
Original post by MatthewAV8B
Questions are good , its how we learn.
Yeah because its greater than , we leave the value as the Raw 1 - whatever , because its not tabulated if its less than . Its hard to explain as I only ever did stats 1 , so I dont know all the detail like a teacher would.
If it was less than , we would need to do one minus our already one misused value if you understand , sorry this bit isnt as clear as the rest :P , ill look it up in the book for further info if you need.
All the best.
Yeah because its greater than , we leave the value as the Raw 1 - whatever , because its not tabulated if its less than . Its hard to explain as I only ever did stats 1 , so I dont know all the detail like a teacher would.
If it was less than , we would need to do one minus our already one misused value if you understand , sorry this bit isnt as clear as the rest :P , ill look it up in the book for further info if you need.
All the best.
No it’s fine , thank you , can I ask u questions if I have any in the near future ? Because I have an upcoming exam soon
Original post by Universecolors
No it’s fine , thank you , can I ask u questions if I have any in the near future ? Because I have an upcoming exam soon
Of course! Just send me a message
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