Statistics(math)

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(20.47 S-AQ) A sample survey interviews an SRS of 252 college women. Suppose (as is roughly true) that 71% of all college women have been on a diet within the past 12 months.

Using Normal approximation, what is the probability (±0.001) that 77% or more of the women in the sample have been on a diet? .

  • Statistics(math) -

    n = 252
    phat = 0.77
    p = 0.71
    q = 1-p = 0.29
    standard deviation = sqrt((pq/n))

    = sqrt(0.7)(0.29)/252)

    = 0.02858

    z score = (phat -p)sd/sqrt(n)

    z = (0.77-0.71)/0.02858

    z = 2.10

    P(phat > 0.77) = p(z < 2.10)

    1- 0.9821 = 0.0179

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