Statistics 1: Post Your Doubts Here

[tanmaydube]

Please someone help is required to solve this out

6 (i) A manufacturer of biscuits produces 3 times as many cream ones as chocolate ones. Biscuits are
chosen randomly and packed into boxes of 10. Find the probability that a box contains equal
numbers of cream biscuits and chocolate biscuits.


(ii) A random sample of 8 boxes is taken. Find the probability that exactly 1 of them contains equal
numbers of cream biscuits and chocolate biscuits.


(iii) A large box of randomly chosen biscuits contains 120 biscuits. Using a suitable approximation,
find the probability that it contains fewer than 35 chocolate biscuits.


This is from Nov 2002 P6 (S1)

Thanks.

3 times as many cream ones as chocolate ones therefore if chocolate is x then cream will be 3x so x+3x=1, 4x=1, x=1/4 and 3x =3/4
Using Binomial we get 10C5*(0.25)^5*(0.75)^5=0.0584

8C1*(0.0584)(1-0.0584)^7=0.307

mean=npq=120*0.25=30
variance=npq=120*0.25*0.75=22.5

P(X<35)~~~~ continuity correction we get P(X<34.5)
P(Z<(34.5-35)/sqr22.5))=P(Z<0.949)
=0.829

 

[Amy Bloom]

Can anybody answer my doubts on pg 2?

 

[bamteck]

3 times as many cream ones as chocolate ones therefore if chocolate is x then cream will be 3x so x+3x=1, 4x=1, x=1/4 and 3x =3/4
Using Binomial we get 10C5*(0.25)^5*(0.75)^5=0.0584

8C1*(0.0584)(1-0.0584)^7=0.307

mean=npq=120*0.25=30
variance=npq=120*0.25*0.75=22.5

P(X<35)~~~~ continuity correction we get P(X<34.5)
P(Z<(34.5-35)/sqr22.5))=P(Z<0.949)
=0.829

OMG ! Thanks a lot. You are great ! Thank you very much.
Btw are u giving for Oct/Nov 2012 exams?

 

[tanmaydube]

Can anybody answer my doubts on pg 2?

which one?

 

[tanmaydube]

OMG ! Thanks a lot. You are great ! Thank you very much.
Btw are u giving for Oct/Nov 2012 exams?

your welcome! yes giving in OCT/NOV 2012 are you giving English Language?

 

[bamteck]

Me too on OCT/NOV 2012 !
Actually I'm giving for General Paper (8009)
Do you do Maths,Chemistry & Physics as well ?

 

[tanmaydube]

Me too on OCT/NOV 2012 !
Actually I'm giving for General Paper (8009)
Do you do Maths,Chemistry & Physics as well ?

yes giving chemistry, english language and mathematics! gave physics in may/june!

 

[soumayya]

Amy Bloom :
Ending in 2: _ _ _ _ _ _ 2 = 6!
Ending in 6: _ _ _ _ _ _ 6 = 6!/2
Ending in 8: _ _ _ _ _ _ 8 = 6!/2

Ans = 6! + 6!/2 + 6!/2 =1440

 

[Amy Bloom]

which one?

This one:

 

[Amy Bloom]

Amy Bloom :
Ending in 2: _ _ _ _ _ _ 2 = 6!
Ending in 6: _ _ _ _ _ _ 6 = 6!/2
Ending in 8: _ _ _ _ _ _ 8 = 6!/2

Ans = 6! + 6!/2 + 6!/2 =1440

Can you remind me which question u're answering?

 

[bamteck]

This one:

View attachment 14811

This one is tough

 

[Iadmireblue]

This one:

View attachment 14811

for question 7 (i)
"if each friend is standing next to his partner"
So If there are 7 couples,they can each stand as BG or GB,and there are 7 "BG or GB"
For the arrangement,it's 7! because any couple can stand first * 2! * 2! * 2!.....7 times because the couple can be BG or GB
which makes the final answer,7! * (2!)^7

for part ii
The Men can stand at 7! ways and the women can stand at 7! ways,and it can be all women and men or all men and women
so the answer is 7! * 7! * 2
Hope it helped

 

[Amy Bloom]

Yup it did. Thanks ^_^

for question 7 (i)
"if each friend is standing next to his partner"
So If there are 7 couples,they can each stand as BG or GB,and there are 7 "BG or GB"
For the arrangement,it's 7! because any couple can stand first * 2! * 2! * 2!.....7 times because the couple can be BG or GB
which makes the final answer,7! * (2!)^7

for part ii
The Men can stand at 7! ways and the women can stand at 7! ways,and it can be all women and men or all men and women
so the answer is 7! * 7! * 2
Hope it helped

 

[Amy Bloom]

Help nightrider1993

 

[celoth]

Help
View attachment 15050

let me try,
1st part...

the Z-value for 0.9 is 1.281,
P(X>5.2)=0.9
P(Z>-1.281)=0.9

therefore,.
-1.281= (5.2 - 2x) / x (x represent standard deviation since i don't know how to make the std deviation symbol appear.)
x = 7.23


2nd part...

Let m represent mean, s represent standard deviation

to know from (m-s) to (m+s), it took how many area,
we will search for this:
P( m-s < X < m+s)
= P ((m-s-m)/s < Z < (m+s-m)/s) )
= P (-1<Z<1)
= 0.8413-0.1587
= 0.6826

then 800 x 0.6826 = 546.08

so we will expect 546.08 observations are between that range.

 

[Amy Bloom]

Help guys!

 

[bamteck]

www.xtremepapers.com/papers/CIE/.../9709_s04_qp_6.pdf
HELP NEEDED for no. 5

 

[Khalifa]

www.xtremepapers.com/papers/CIE/.../9709_s04_qp_6.pdf
HELP NEEDED for no. 5

link isnt working

 

[Khalifa]

Help guys!
View attachment 17688

Q3-Probability Of scoring a 12:

You get odd number on the spinner: 3/5
You add the dices, you get 12 only from 6+6
So P(12)= 1/36

As they are succesive events p(12)= (3/5)*(1/36)
= 1/60

You get an even number on the spinner: 2/5
You multipy the dices- you get 12 in 4 ways
----2 *6, 6*2, 3*4, 4*3
P(12) = 4/36

As they are successive events as well: (2/5)*(4/36)
=2/45

To get the total probability of getting a 12, add the 2 above Ps as the two events abover are mutually exclusive:
2/45 + 1/60= 11/180

Now tp find out the conditional probablity
P(Even no.| get 12) = P(even no. Intersection 12) / P(12)

=(2/45)/(11/180)
=8/11


Q.5 I found this one easier

Suppose x= 320 + d

P(X<x)
P(Z< x-320/21.6)= 0.409

Let phi=¥

¥(x-320/21.6)= .409
(x-320/21.6)= ¥^-1(.409)

As .409 cant be read off the table, you will take negative value

X-320/21.6= -¥^-1(0.591)

Read off the table

X-320/21.6=- .230
X=324.968

X= 320+d
d= 4.968
~ 4.97

 

[babycry]

Help guys!
View attachment 17688

for 5ii an easier way to think of it is:

since it is at least 320+d, it implies p(z>320+d) since it is at least. then this gives a z value of 0.23 when read off the table. then do 320+d-320/21.6=0.23 to give d to be 4.97