Statistics Question Please Help Asap!!?

A certain flight arrives on time 81 percent of the time. Suppose 197 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly 166 flights are on time.

​(b) at least 166 flights are on time.

​(c) fewer than 169 flights are on time.

​(d) between 169 and 175​, inclusive are on time

1 Answer

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  • fcas80
    Lv 7
    4 weeks ago

    Normal approximation to binomial

    n =  197

    p  = 0.81

    xbar = np  = 159.57

    s = √[n * p * (1-p)] =  5.51

    I am going to use the continuity correction factor.  Some books and teachers use it, and some do not.  If yours does not, then the following may be wrong.

    x = 166

    x - .5  = 165.5

    x + .5 =  166.5

    z for x - .5 = 165.5 is 1.08

    P(z < 1.08) = 0.8599

    z for x + .5 = 166.5 is  1.26

    P(z < 1.26) = 0.8962

    a.  P(x=166) = 0.8962 - 0.8599 = 0.0363

    b.  P(x>=166) = P(x>=165.5) = 1 - .8962 = .1038

    c.  P(x < 169) = P(x < 168.5) = P(z < 1.62) = 0.9474

    P(x < 175) = P(x < 174.5) = P(z < 1.62) = 0.9966

    d.  P(169 < x < 175) = 0.0492

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