Posted by Robbie on .
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,
Z = (scoremean)/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,
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 Zscore 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.