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A normal distribution has a mean of 100 and standard deviation of 20. What is the
probability of randomly selecting a score less than 130 from this distribution?
a, p=0.9032
b, p=0.9332
c, p=0.0968
d, p=0.0668

  • statistics -

    Z = (score-mean)/SD

    Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to the Z score.

  • statistics -

    First notice they gave you the mean, standard deviation, and an X value. Meaning you can you the formula

    I would also draw a distribution picture where the number will end up being.

    X = Z * (standard deviation) + (mean)

    Plug in the numbers:
    130 = Z (20) + 100

    Subtract 100 FROM 130:
    30 = Z (20)


    Your Z-score is:
    Z = 1.5

    *** Since you are looking for a probability LESS than 130, the probability will be on the LEFT side of the distribution. This means you have to look on your "Unit Normal Table" for a Z score of 1.5 and in the "B" (proportion body) column.

    The answer is 0.9332

    **Note every table may vary, so look to see what you B column says for the Z score of 1.5.

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