Question

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

Answer 1

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.

Answer 2

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.