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    The number of telephone calls per day follows a Poisson distribution with mean 2

    Use an approximate normal distribution to estimate the probability of at least 17 calls in in 8 days.

    Newton.
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    X
    = Number of calls in 8 days
    ~ Po(16)

    So X ~ N(16, 16) approximately.

    P(X >= 17)
    ~= P(N(16, 16) >= 16.5) . . . . . continuity correction
    = 1 - Phi(0.5/4)
    = 0.4503
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    I got the same answer as the book arrived at as P(N(0,1)>0.125)
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    Why phi(0.5/4)? Should it not be Phi((1/4)(17.5-16))?
    Yes, if we wanted P(X > 17), ie, P(X >= 18). But the question asks for the probability of "at least 17 calls" not "more than 17 calls"
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    (Original post by Newton)
    Why does P(X>17)=P(X>=18)?

    Newton.
    P(X>17) = P(X>=18) for a discrete integer-valued variable
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    X counts the number of calls so it has to be an integer.

    P(X > 17)
    = P(17 < X < 18) + P(X >= 18)
    = 0 + P(X >= 18)
    = P(X >= 18)
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    (Original post by Newton)
    Ok. Please tell Me if my understanding is then correct.

    It is asking for the probability of at least 17 calls i. e. P(X>=17).

    By continuity correction this transforms to P(X>=17.5).

    But since X must take an integer value we are looking for P(X>=18).

    Newton.
    No we end up working out

    P(N(16,16)>16.5)

    because each interval

    P( n-1/2 < N(16,16) < n+1/2)

    is a good approximation for P(X=n)
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    (Original post by Newton)
    So you minus a half?

    Newton.
    yes - well as appropriate

    if we'd been asked what is P(X <= 17) we'd have worked out P(N(16,16)<17.5)
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    (Original post by Newton)
    But that is what we are being asked for, hence the 'at least'.

    Newton.
    :confused:

    no that would be phrased as "at most 17"

    just remember

    P(X=n) is approximately P(n-1/2 < N(16,16) < n+1/2)

    and add up whichever of these you need
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    (Original post by Newton)
    I mean P(X>=17).

    Newton.
    yes so you need to add

    P(X=17)+P(X=18)+P(X=19)+...

    and by the approx I gave you this is

    P(16.5<N<17.5)+P(17.5<N<18.5)+.. .

    = P(16.5<N)
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    (Original post by Newton)
    tell Me if my understanding is then correct
    why Me?
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    (Original post by yazan_l)
    why Me?
    Obsessive Compulsive Behaviour.

    Newton.
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    /Ok. Please tell Me if my understanding is then correct.

    It is asking for the probability of at least 17 calls i. e. P(X>=17).

    By continuity correction this transforms to P(X>=17.5).

    But since X must take an integer value we are looking for P(X>=18).

    Newton./

    If its P(X>=17) it must include the integer 17 therefore with the continuity correction it becomes P(X>=16.5).

    Just remember what integers you have to include!
 
 
 
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