[lburbank989]

Suppose that a student's verbal score X can be considered an observation from normal distribution having mean 500 and standard deviation 80.
a) Find P(X>660)
b) Find the 90 percentile of the distribution
c) Find the probability that the student scores below 420
d) Find the probability that the student's score between 400 and 600

I am not sure how to start with this question. Any help would be appreciated. I have a few more like it and using this as an example would be great thanks.

[terryg752]

A. To Standardize 660:

Z = (660 - 500)/80 = 2

Look in the Normal Dist table (giving Cumulative probabilities) for z = 2

Probabilty of less than 660 = .9772

Probabilty of > 660 = 1 - .9772 = 0.0228


B. In the percentile table of Normal Dist.

90 Percentile value of Z = 1.282

(X - 500)/80 = 1.282

X = 500 + 80 times 1.282

= 602.56

C Prabability of Less than 420

Corresponding Z = (420 - 500)/80 = -1

Prob. Z < -1 = Prob Z > 1 (because of symmetry of the distribution)

Now proceed like part A.

D. Standardize 400 and 600 as above. Then use

Prob = Prob (X < 600) - Prob (X < 400)