Stat Help

[psycoist14]

O/N 10 61, number 3 i) how do we find that number in integer..?? i found the probability.. 0958 now???

 

[psycoist14]

haha i found the answer.. lol thanks XPF

 

[psycoist14]

http://www.xtremepapers.me/CIE/Internat ... _ms_61.pdf
5 i explanation required please.. i understood the table though didn't get the calculation. :S

 

[samy]

HELP HELP!!

wen do v take standard frequencies while making histograms???

 

[psycoist14]

i didn't understand you.. please tell more briefly and if you can please solve my problem too

 

[Amna]

http://www.xtremepapers.me/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_w10_ms_61.pdf
5 i explanation required please.. i understood the table though didn't get the calculation. :S

okay, so the question asks for the probability that AT LEAST 2 friends choose entrance B. that means either 2 or all 3 of them choose entrance B. for 2 friends, the combinations can be:
B B other
Rick Brenda Ali
Ali Rick Brenda
Brenda Ali Rick

so now, just find the values from the table, plug them in to find the probability for each row. add those together with the probability that all 3 come through entrance B, and there you have the answer.

HELP HELP!!

wen do v take standard frequencies while making histograms???

when making histograms of data that has unequal class width. in such cases you have the option to either draw the histogram using frequency density, or using standards. standards are usually used when all the class widths are multiples of one. for example, if the class widths are 10, 5, 30, 15, 20 respectively, you can use 5 units= 1 standard. so the heights of the bars in the histogram will be obtained by dividing the frequency with 2, 1, 6, 3, 4 respectively. does that make sense? and if you want to use frequency density, you can divide the frequency by the class width to obtain the height of the bar. frequency density is usually the fool-proof method that i prefer... it's just plugging in numbers without using any common sense =P

 

[psycoist14]

i still didnot get what you meant

 

[Amna]

you get the part where at least two friends means either 2 or all 3 friends use the same entrance B? so P(at least 2 friends)= P(2 friends) + P(3 friends)
we know the three friends are Ali, Brenda, and Rick.
so if all three go through entrance B, P= 2/9 x 1/4 x 2/35= 1/315
if two go through entrance B, and one goes through some other entrance, the possible combinations are:
Ali and Brenda use entrance B, and rick goes through some other entrance: P= 2/35 x 1/4 x (1-2/9)= 1/90
Ali and Rick use entrance B, Brenda goes through some other entrance: P= 2/35 x 2/9 x (1-1/4)= 1/105
Rick and Brenda use entrance B, Ali goes through some other entrance: P= 2/9 x 1/4 x (1-2/35)= 11/210
so P(2 friends)= 1/90 + 1/105 +11/210 = 23/315
so P(at least 2)= 23/315 + 1/315 = 24/315 = 8/105.
now do you get it?
if not can you tell me a particular part of the explanation that you don't get?

 

[psycoist14]

superbly explained... i've now become crystal clear thanks mate..

 

[Amna]

lol welks =P