Random variable X distributes normally. 10% of the distribution is above 125, 20% is below 77. What is the mean and variance of this distribution?

Answer 1

As X has a normal distribution, if we are given the probability that X is greater than a value or less than a value, we can determine the z value. Thus, we can determine the z value at 125 and the z-value at 77. The difference in z-values tells us how many standard deviations 77 is from 125. We can then determine the standard deviation. Finally, by going the standard deviation and the z value at 125, we can determine the mean.

If P(X >= 125) = .1, P(X <= 125) = ..9 (X has a continuous distribution, so we don't have to worry about <= versus <.)

From Excel, if P(X <= 125) = .9, z at 125 is norm.s.inv(.9) = 1.2815515655446

P(X <= 77) = .2, so z at 77 = norm.s.inv(.2) = -0.841621233572915

Then 77 and 125 are 1.2815515655446 - -0.841621233572915 = 2.12317279911751 standard deviations apart.

Thus, the standard deviation is (125-77)/2.12317279911751 = 22.607674712087, and the variance is the standard deviation squared = 22.607674712087 ² = 511.106955887538

Using the z value of 125 and the standard deviation, the mean is

125 - 1.2815515655446 * 22.607674712087 = 96.0270990794018

mean is 96.0270990794018

variance is 511.106955887538