Mathematics Stats Paper 6 Help

[HubbaBubba]

1. The volume of milk in millilitres in cartons is normally distributed with mean (u) and standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it was found that 225 of them contained more than 1002 millilitres.
(i) Calculate the value of (u)

Can anybody tell me how they found the value of z ?

 

[Fadhila]

1. The volume of milk in millilitres in cartons is normally distributed with mean (u) and standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it was found that 225 of them contained more than 1002 millilitres.
(i) Calculate the value of (u)

Can anybody tell me how they found the value of z ?

Ur probability is 225/900=0.25
p(x>1002)=0.25
p(x<1002)=0.75
p(z<0.674)=0.75

 

[kiara15]

i had some ambiguity in below quest:
1) P(Z>a)=0.8888 , a is negativ,
1-fi(-a)=o.8888
1-[1-fi(a) ]=o.8888
the other solution of quest i found:
2)p(z<t)=0.25, t is negativ
fi(-t)=0.25
1-fi(t)=0.25..
can anyone tell me this difference and concept how to solve such type of quest plzzz

 

[pakiboy]

I am little bit cnfusd with the boundaries and class width thng can any one here help me regarding ths ?

 

[Fadhila]

I am little bit cnfusd with the boundaries and class width thng can any one here help me regarding ths ?

You usually get class width and stuff to draw a histogram what u need is boundries which must be continuous. ur class width is the diffrence of ur boundries ( For ex: 0.5-5.5, ur class width is 5.5-0.5)
frequency density= frequency/ class width
Hope it helps

 

[Fadhila]

i had some ambiguity in below quest:
1) P(Z>a)=0.8888 , a is negativ,
1-fi(-a)=o.8888
1-[1-fi(a) ]=o.8888
the other solution of quest i found:
2)p(z<t)=0.25, t is negativ
fi(-t)=0.25
1-fi(t)=0.25..
can anyone tell me this difference and concept how to solve such type of quest plzzz

I dont understand the question

 

[panoramafolks]

i had some ambiguity in below quest:
1) P(Z>a)=0.8888 , a is negativ,
1-fi(-a)=o.8888
1-[1-fi(a) ]=o.8888
the other solution of quest i found:
2)p(z<t)=0.25, t is negativ
fi(-t)=0.25
1-fi(t)=0.25..
can anyone tell me this difference and concept how to solve such type of quest plzzz

can u post the particular question??

 

[pakiboy]

You usually get class width and stuff to draw a histogram what u need is boundries which must be continuous. ur class width is the diffrence of ur boundries ( For ex: 0.5-5.5, ur class width is 5.5-0.5)
frequency density= frequency/ class width
Hope it helps

I faced prob. While solvng cumulative freq. Quest. Especially when finding the mean!
I just want to knw when u add 0.5 in class boundaries

 

[Fadhila]

I faced prob. While solvng cumulative freq. Quest. Especially when finding the mean!
I just want to knw when u add 0.5 in class boundaries

You add and deduct 0.5 when ur class boundries are nt continuous.. Check the 6th question-http://papers.xtremepapers.com/CIE/...S Level/Mathematics (9709)/9709_w09_qp_62.pdf
Since the boundries are not continuous u take 0.5-20.5, then next one 20.5-30.5, next 30.5-40.5 and so on.. So for the 1st value u minus 0.5 and the second value add 0.5.
When it comes to the mean in grouped frequency: take the midvalue of the class boundry,and in the case of this 6th question, that is done by adding (1+21)/2 * ur frequency and do it in the same manner add ur answrs and divide by the total!

 

[syed1995]

You add and deduct 0.5 when ur class boundries are nt continuous.. Check the 6th question-http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_62.pdf
Since the boundries are not continuous u take 0.5-20.5, then next one 20.5-30.5, next 30.5-40.5 and so on.. So for the 1st value u minus 0.5 and the second value add 0.5.
When it comes to the mean in grouped frequency: take the midvalue of the class boundry,and in the case of this 6th question, that is done by adding (1+21)/2 * ur frequency and do it in the same manner add ur answrs and divide by the total!

Thanks man.. that clears some concepts...

 

[pakiboy]

You add and deduct 0.5 when ur class boundries are nt continuous.. Check the 6th question-http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Mathematics (9709)/9709_w09_qp_62.pdf
Since the boundries are not continuous u take 0.5-20.5, then next one 20.5-30.5, next 30.5-40.5 and so on.. So for the 1st value u minus 0.5 and the second value add 0.5.
When it comes to the mean in grouped frequency: take the midvalue of the class boundry,and in the case of this 6th question, that is done by adding (1+21)/2 * ur frequency and do it in the same manner add ur answrs and divide by the total!

thnk u soo much i got it
u have definitely saved 3 to 4 marks for me!

 

[Fadhila]

No problem

 

[salvatore]

Find how many different odd numbers greater than 500 can be made using some or all of the digits 1, 3, 5 and 6 with no digit being repeated.

I need help with this question.. could someone please explain it to me?
Thanks

 

[Kangxidi]

odd numbers greater than 500.

Lets count how much we can do in 3 digits.

500-999.

> 500 therefore 1st digit should be either 5 or 6.

5 _ _ OR 6 _ _

5 _ _ = 3P2 - 2 [-2 is because 6 cannot be the end number or it will be even, therefore subtracting two possible arrangements of 1 and 6 + 3 and 6]

6 _ _ = 3P2

Therefore between 500-999 there are (3P2 - 2) + 3P2 ways.

Now consider > 1000

We can have ANYTHING, apart from 6 being at the back.

Therefore let's count the number of ways in which the number 1,3, 5 and 6 can be arranged MINUS the number of ways in which number 6 would be at the end.

Number of ways in which 1 3 5 6 can be arranged, will be 4!

Number of ways in which 6 will be at the back = 3!

Therefore in > 1000, there are 4! - 3! possible combinations of odd numbers with number 1 3 5 and 6.

In the end, add them together.

(3P2 - 2) + 3P2 + (4! - 3!)

Work it out with your calculator I don't have mine around atm if the answer is incorrect tell me, I must have missed some stuffs myself (I am not expert in permutations/combinations myself, just trying to help )

 

[salvatore]

5 _ _ = 3P2 - 2 [-2 is because 6 cannot be the end number or it will be even, therefore subtracting two possible arrangements of 1 and 6 + 3 and 6]

Thanks a lot.. your answer is correct!

I still don't get why you subtracted 2.. as in, how did you get two arrangements?

 

[Kangxidi]

Thanks a lot.. your answer is correct!

I still don't get why you subtracted 2.. as in, how did you get two arrangements?

Let's assume we are writing odd numbers between 500 to 600. This automatically force us to put 5 as the first number.

We will have 5 _ _ (Two empty space, in which we can put 1,3 OR 6 in any of the 2 empty spaces)

Consider all the arrangements of 3P2 (three numbers to be inserted to two space), there will be 6 different arrangements, namely:

• 13
• 16
• 31
• 36
• 61
• 63

Since we want an ODD combination, the combination of 16 and 36 highlighted above cannot be a part of the appropriate arrangement, as inserting 16 and 36 to your two spots above ^ would make the number 516 and 536 respectively, which are even numbers. Therefore we subtract those two arrangements, leaving us with only 4 appropriate ones, 13,31,61 and 63.

 

[salvatore]

Let's assume we are writing odd numbers between 500 to 600. This automatically force us to put 5 as the first number.

We will have 5 _ _ (Two empty space, in which we can put 1,3 OR 6 in any of the 2 empty spaces)

Consider all the arrangements of 3P2 (three numbers to be inserted to two space), there will be 6 different arrangements, namely:

• 13
• 16
• 31
• 36
• 61
• 63

Since we want an ODD combination, the combination of 16 and 36 highlighted above cannot be a part of the appropriate arrangement, as inserting 16 and 36 to your two spots above ^ would make the number 516 and 536 respectively, which are even numbers. Therefore we subtract those two arrangements, leaving us with only 4 appropriate ones, 13,31,61 and 63.

Oh yeah.. got it!

Thank you