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Help with statistics question PLEASE?

An airline conducts the following research. Average evacuation time in the past has been 100 seconds with a standard deviation of 15 seconds.

If the airline conducts a sample of 30 tests, what is the probability that the average evacuation time will be 97 seconds or less?

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  • 1 decade ago
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    the std dev of sample means (ss) is std dev of pop/sqrt(sample size)

    ss= 15/sqrt(30)=15/5.48=2.73

    z=(xbar-mean)/sd

    change 97 to a z-score

    z=(97-100)/2.73

    z=-3/2.73=-1.10

    z=-1.10

    central area for z=-1.10 is .3643

    area in question is to LEFT of that area (tail end)

    subtract from .5

    .5000-.3643=.1357 (answer)

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