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    It is part C in which I am struggling with - I have looked on the tables for p(x<= 10) for the last value in which it is greater than 0.9, I found this value, but then the final answer on the markscheme correlates to a different value?

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    (Original post by iMacJack)
    It is part C in which I am struggling with - I have looked on the tables for p(x<= 10) for the last value in which it is greater than 0.9, I found this value, but then the final answer on the markscheme correlates to a different value?

    Thank you
    Look at the percentage points table, i.e: the table right below the normal distribution one in your formula booklet, you'll see that it gives a z value of 1.2816.
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    (Original post by Zacken)
    Look at the percentage points table, i.e: the table right below the normal distribution one in your formula booklet, you'll see that it gives a z value of 1.2816.
    I don't really see how this is helping me
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    (Original post by iMacJack)
    I don't really see how this is helping me
    Oh, sorry. I assumed it was a normal distribution question without looking at the previous parts.

    \mathbb{P}(C &gt; 10) &lt; 0.1 \iff \mathbb{P}(C \leq 10) &gt; 0.9. Now look in the x=10 row of your poisson table.
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    (Original post by Zacken)
    Oh, sorry. I assumed it was a normal distribution question without looking at the previous parts.

    \mathbb{P}(C &gt; 10) &lt; 0.1 \iff \mathbb{P}(C \leq 10) &gt; 0.9.
    According to the markscheme, apparently not
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    (Original post by iMacJack)
    According to the markscheme, apparently not
    Does this help:
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    (Original post by Zacken)
    Does this help:
    Yeah - they were the two values I found, but then in the markscheme the final answer is lambda = 7/4?
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    (Original post by iMacJack)
    Yeah - they were the two values I found, but then in the markscheme the final answer is lambda = 7/4?
    Yes, because Charlotte does it over 4 hours. So, really what you've found is C \sim Po(\mu) where \mu = 7.

    But the random variable X \sim Po(\lambda) is over 1 hour. So C \sim Po(4\lambda), that is: \mu = 4\lambda.
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    (Original post by Zacken)
    Yes, because Charlotte does it over 4 hours. So, really what you've found is C \sim Po(\mu) where \mu = 7.

    But the random variable X \sim Po(\lambda) is over 1 hour. So C \sim Po(4\lambda), that is: \mu = 4\lambda.
    Oh yeah... should've realised that

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    (Original post by iMacJack)
    Oh yeah... should've realised that

    Thank you
    No problem.
 
 
 
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