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AQA M1SB 8th of june 2016 unofficial markscheme

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Binomial

ai) Probability there were exactly 4 red loom bands

(50C4) * (0.18)^4 * (0.82)^46 = 0.0262

aii) Calculate the probability at most 10 were yellow

Therefore X ~ B(50, 0.15)

P(X≤10) = 0.8801

aiii) Something to do with green and purple bands which gave you

X ~ B(50, 0.5)


aiv) Probability that more than 35 but less than 45 bands were neither yellow nor white

Thus you have X ~ B(50, 0.8)

For P(35 < X < 45)

Therefore using the complementary event denoted by Y

Y ~ B(50, 0.2)

P(6 ≤ Y ≤ 14) = P(Y≤14) – P(Y≤5) = 0.8913

b) Calculate the mean and variance for 300 red loom bands

μ = np → 300 * 0.18 = 54

σ² = np(1–p) → 54(1–0.18) = 44.3
Original post by Chickenslayer69
Yeah, I know that I did it wrong and I know what I did wrong, but I don't know how many marks I'd get for forgetting to divide by square root of n. I'd definitely lose 2 for getting the interval wrong, but I'm not sure about the other 2...

Not to crush hopes but I think there's a mark for getting the standard error.
Original post by CirclesAreRound
Binomial

ai) Probability there were exactly 4 red loom bands

(50C4) * (0.18)^4 * (0.82)^46 = 0.0262

aii) Calculate the probability at most 10 were yellow

Therefore X ~ B(50, 0.15)

P(X≤10) = 0.8801

aiii) Something to do with green and purple bands which gave you

X ~ B(50, 0.5)


aiv) Probability that more than 35 but less than 45 bands were neither yellow nor white

Thus you have X ~ B(50, 0.8)

For P(35 < X < 45)

Therefore using the complementary event denoted by Y

Y ~ B(50, 0.2)

P(6 ≤ Y ≤ 14) = P(Y≤14) – P(Y≤5) = 0.8913

b) Calculate the mean and variance for 300 red loom bands

μ = np → 300 * 0.18 = 54

σ² = np(1–p) → 54(1–0.18) = 44.3

Yay that's what I got :wink:
What was the last probability question even on about though?

It said 4 people *not* from the sample of 500 so I was completely stumped as to where to get my probabilities from :-\

I know that there were six combinations possible so the end answer had to be multiplied by 6/3! but apart from that colour me confused.

I used the probabilities from the table regardless because I had no other option I could think of
Reply 44
Avatar for SM-
SM-
6
For the last part of the table one I got 0.00916 and or the spring water bottles I assumed the bottles volumes were independent of each other, that right? Also for that part is it ask you to say whether each claim is likely or unlikely?


The mean was reduced by 12.4?
(edited 7 years ago)
For the last part of the normal distribution question, did anyone else say the mean would have had to have reduce by about 18.2? (can;t remember exact)
Original post by CirclesAreRound
What was the last probability question even on about though?

It said 4 people *not* from the sample of 500 so I was completely stumped as to where to get my probabilities from :-\

I know that there were six combinations possible so the end answer had to be multiplied by 6/3! but apart from that colour me confused.

I used the probabilities from the table regardless because I had no other option I could think of

I also used the probablities from the table but I can't remember what answer I got
Original post by jake4198
For the last part of the normal distribution question, did anyone else say the mean would have had to have reduce by about 18.2? (can;t remember exact)

I said it reduced by 2.3
Reply 48
Avatar for JPencil
JPencil
9
Anyone else leave all of the probabilities in fractional form...? :frown:

Posted from TSR Mobile
Original post by CirclesAreRound
What was the last probability question even on about though?

It said 4 people *not* from the sample of 500 so I was completely stumped as to where to get my probabilities from :-\

I know that there were six combinations possible so the end answer had to be multiplied by 6/3! but apart from that colour me confused.

I used the probabilities from the table regardless because I had no other option I could think of


You did need to use the probabilites from the table, as it wanted you to use the sample to estimate the probability of events occuring in the population.
Original post by JPencil
Anyone else leave all of the probabilities in fractional form...? :frown:

Posted from TSR Mobile

... no it said to 3 decimal places
Original post by JPencil
Anyone else leave all of the probabilities in fractional form...? :frown:

Posted from TSR Mobile


From previous past papers you'd be penalised a maximum of an A1 mark once only so it shouldn't be too bad!
Original post by Dapperblook22
You did need to use the probabilites from the table, as it wanted you to use the sample to estimate the probability of events occuring in the population.


In that
Original post by CirclesAreRound
In that


In that you were using the sample of people to estimate the outcome for 4 people from the whole surgery.
Original post by Dapperblook22
In that you were using the sample of people to estimate the outcome for 4 people from the whole surgery.


No sorry my phone died! I meant to say:

"In that case *cheers* :-D"

Sorry for the confusion!
Original post by CirclesAreRound
No sorry my phone died! I meant to say:

"In that case *cheers* :-D"

Sorry for the confusion!


Ah, ok :tongue:. No worries.
Reply 56
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SM-
6
What probability did you guys use for the table one 148 over 500 and 66 over 500?
What assumption did you guys make for the spring water one? Did they question say to put likely or unlikely for each part?
Was the mean reduced by 12.4?
Did anyone else get 0.8913 for question 5b?
I got the first claim as valid , what did other people get ?
I'm really confused about the reduction question. I got 1.7.

1535-u/9.6=1.28 (90%)

U= -(9,6(1.28)-1535)

1525 - u gives reduction right?

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