Posted by **Robbie** on Wednesday, November 7, 2012 at 8:48pm.

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 -
**PsyDAG**, Thursday, November 8, 2012 at 10:53am
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 -
**Kristin**, Tuesday, April 12, 2016 at 2:34pm
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)

Divide:

30/20

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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