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A sample survey interviews an SRS of 263 college women. Suppose (as is roughly true) that 75% of all college women have been on a diet within the past 12 months.



Using the Normal approximation, what is the probability (±0.001) that 82% or more of the women in the sample have been on a diet?

  • statistics -

    Use the binomial approximation to a normal distribution.

    mean = np = 263 * .75 = ?
    standard deviation = √npq = (√(263 * .75 * .25) = ?
    Note: q = 1 - p
    Finish the calculation.

    Next, use z-scores:

    z = (x - mean)/sd

    x = .82
    Use mean and sd calculated above.

    Finally, use a z-table to determine your probability (remember the question is asking "82% or more" when looking at the table).

    I hope this will help get you started.

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