by dud on February 9th, 2011

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On a normal curve table, what percentage of scores fall between a Z score of 1.29 and a Z score of 1.49? (on a normal curve)

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  • by engineer is Terminator on February 9th, 2011
    voted: 3.04

    engineer is Terminator

    You can't do this without the use of z score table.
    http://www.statsoft.com/textbook/distribution-tables/

    From the table, you see for 1.49 the z score is 0.4319

    For 1.29 the Z score is 0.4015

    So the area enclosed = 0.4319 - 0.4015 =0.0304

    Hence the % of area enclosed is 0.0304 * 100 = 3.04% ans

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