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24 March, 14:00

Suppose that 76% of americans prefer coke to pepsi. A sample of 200 was taken. What is the probability that at least sixty eight percent of the sample prefers coke to pepsi?

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  1. 24 March, 14:36
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    0.996

    Step-by-step explanation:

    It is given 76% of Americans prefer Coke to Pepsi.

    A sample of 200 was taken, i. e n=200

    Let X be the random variable denoting the number of Americans that prefer Coke to Pepsi.

    Then X follows a Binomial Distribution with p=0.76

    To find P (X>=136) [68% of 200 = 136]

    P (X>=136) = 1-P (X<136) = 1-P (X<=135)

    =1-0.003979

    =0.996020
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