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Math Help - request for normal distribution exercise

  1. #1
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    request for normal distribution exercise

    I am kindly asking for your help considering the following exercise.


    The probability to win in a roulette game is 18/37. If someone bets 100 times, which is the probability to win at least 50 times?
    (an excellent approximation to B(n, p) is given by the normal distribution, where μ=np and σ2=np(1-p))


    Solution

    P (50 ≤ x ≤ 100) = P ((50-μ)/σ ≤( x-μ )/σ≤( 100-μ)/σ) = Φ((100-μ)/σ) – Φ((50-μ)/σ)

    [ where μ=np, σ2=np(1-p), n=100, p=18/37 ]

    then we use the normal distribution table
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  2. #2
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    You are on the right track. Consider where X\text{~ Bi}(n,p) and Y\text{~ N}(\mu,\sigma) then P(X\leq x) \approx P(Y\leq x+0.5)
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  3. #3
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    thanks for your reply, pickslides!

    I figured out a solution. Do you think this is correct?

    p=0,5
    X ~ Bin(100, 0.5)

    X is approximately normally distributed with μ = (n × p) = 100 × 0.5 = 50 and
    σ2 = n × p × (1-p) = 100 × 0.5 × 0.5 = 25
    X is approximately N(50,25)

    P (x≥50) = 1 – P(x < 50) = 1 – P (z<(50-μ)/σ) = 1 - P (z<(50-50)/5) = 1 - P(z<0) = 1 – Φ(0) = 0.5
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