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g Twenty percent of drivers driving between 11 PM and 3 AM are drunken drivers. Using the binomial probability formula, find the probability that in a random sample of 12 drivers driving between ll PM and 3 AM, two to four will be drunken drivers. (Round to 4 digits, ex. 0.1234)

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  1. 14 July, 16:41
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    Answer: p (2 lesser than or equal to x lesser than or equal to 4) = 0.6526

    Step-by-step explanation:

    20% of drivers driving between 11 PM and 3 AM are drunken drivers.

    We want to use the binomial distribution to determine the probability that in a random sample of 12 drivers driving between 11 PM and 3 AM, two to four will be drunken drivers.

    The formula for binomial distribution is

    P (x = r) = nCr * q^n-r * p^r

    x = number of drivers

    p = probability that the drivers that are drunken.

    q = 1-p = probability that the drivers are not drunken.

    n = number of sampled drivers.

    From the information given,

    p = 20/100 = 0.2

    q = 1 - p = 1 - 0.2 = 0.8

    n = 12

    We want to determine

    p (2 lesser than or equal to x lesser than or equal to 4)

    It is equal to p (x=2) + p (x = 3) + p (x=4)

    p (x=2) = 12C2 * 0.8^10 * 0.2^2 = 0.2835

    p (x=3) = 12C3 * 0.8^9 * 0.2^3 = 0.2362

    p (x=4) = 12C4 * 0.8^8 * 0.2^4 = 0.1329

    p (2 lesser than or equal to x lesser than or equal to 4) = 0.2835 + 0.2362 + 0.1329 = 0.6526
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