Question

Suppose that X is normally distributed with mean 95 and standard deviation 13.

A. What is the probability that X is greater than 113.72?
Probability =

B. What value of X does only the top 14% exceed?
X =

Answer 1

Solution:- Given that mean = 95, standard deviation = 13

A. P(X > 113.72) = P((X-mean)/sd > (113.72-95)/13)
= P(Z > 1.44)
= 1 − P(Z < 1.44)
= 1 − 0.9251
= 0.0749
Use the standard normal table to find the area below 1.44 :

| Z | 0.0 0.1 0.2 0.3 0.4 0.5 10.6 0.7 0.8 09 1.0 1.1 12 1.3 1.4 0.00 0.01 0.02 0.03 0.04 05 0.504 0.508 0512 0.516 0.5398 15438 1.5478 0.5517 0.5557 0.5793 15832 0.5871 0.591 0.5948 0.6179 06217 0.6255 0.6293 0.6331 0.6554 0.6531 0.628 0.6664 0.67 0.6915 0.695 0.6985 0.7019 0.7054 0.7257 0.7291 0.7324 0.7357 0.7389 0.758 0.7611 0.7642 0.7673 0.7704 0.7881 0.791 0.7939 0.7967 0.7995 0.8159 0.8186 0.8212 0.3238 0.3264 0.3413 0.8438 0.3461 0.3485 0.3503 0.8643 0.8665 13686 0.3703 0.3729 0.8849 0.8863 1388 0.3907 0.3925 0.9032 0.4049 19066 0.9082 0.9099 09192 19207 13222 0.9236 092511

B. for Z = 1.08

X = mean + Z*Sd
= 95 + 1.08*13
= 109.04

Answer 2

where are you getting 1.08