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Math Help - Help me again, urgent

  1. #1
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    Help me again, urgent

    In a large city 1 person in 5 is left handed.
    i) find probability that in a random sample of 10 people exactly 3 will be left handed
    ii) find the most likely number of left handed people in a random sample of 12 people
    iii) find the mean and standard deviation of the number of left handed people in a random sample of 25 people
    iv) How large must a random sample be if the probability that it contains at least one left handed person is to be greater than 0.95?
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  2. #2
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    Quote Originally Posted by whleow View Post
    In a large city 1 person in 5 is left handed.
    i) find probability that in a random sample of 10 people exactly 3 will be left handed
    ii) find the most likely number of left handed people in a random sample of 12 people
    iii) find the mean and standard deviation of the number of left handed people in a random sample of 25 people
    iv) How large must a random sample be if the probability that it contains at least one left handed person is to be greater than 0.95?
    i) n=10, p=0.2,
    P(X=3)= 10C3x (0.2)^3x (0.8)^7

    = 0.201326592

    ii) mean= E(X) = np = 12x0.2= 2.4 or 2 people.

    iii)mean = E(X) = np = 25x0.2 = 5 people.
    S.D. = (Var(X))^0.5 = (np(1-p))^0.5 = (25x0.2x0.8)^0.5 = 2

    iv) Problem
    I'm stuck here, I got this far:

    1-(P(X=0))=0.95 which goes to

    0.05=(P(X=0))

    0.05=n(1-p)

    0.05=n(0.8)

    n=0.0625...?

    Where did I go wrong?
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  3. #3
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    Quote Originally Posted by Nyoxis View Post
    i) n=10, p=0.2,
    P(X=3)= 10C3x (0.2)^3x (0.8)^7

    = 0.201326592

    ii) mean= E(X) = np = 12x0.2= 2.4 or 2 people.

    iii)mean = E(X) = np = 25x0.2 = 5 people.
    S.D. = (Var(X))^0.5 = (np(1-p))^0.5 = (25x0.2x0.8)^0.5 = 2

    iv) Problem
    I'm stuck here, I got this far:

    1-(P(X=0))=0.95 which goes to

    0.05=(P(X=0))

    0.05=n(1-p)

    0.05=n(0.8)

    n=0.0625...?

    Where did I go wrong?
    The smallest integer value of n satisfying the following is required:

    \Pr(X = 0) \leq 0.05

    \Rightarrow {n \choose 0} p^0 (1 - p)^n \leq 0.05

    \Rightarrow (1) (1) \left( 1 - \frac{1}{5} \right)^n \leq 0.05

    \Rightarrow \left( \frac{4}{5} \right)^n \leq 0.05

    \Rightarrow n = 14.
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