Question

Let y have the lognormal distribution with mean 81.6 and variance 158.90.

Compute the following probabilities P(y < 60)

Answer 1

The probability y is

P(Y > 111) = 0.0329

P(79 < Y < 111) = 0.4622

(a) P(Y > 111)

To calculate this probability, we first need to find the z-score of 111. We can do this by subtracting the mean of the lognormal distribution (81.6) from 111 and then dividing by the standard deviation, which is the square root of the variance (158.90).

z = (111 - 81.6) / sqrt(158.90) = 1.84

We can then look up the probability of z > 1.84 in a z-table. This value is 0.0329.

Therefore, the probability that Y is greater than 111 is 0.0329.

(b) P(79 < Y < 111)

To calculate this probability, we need to find the z-scores of 79 and 111. We can do this using the same method as above.

z1 = (79 - 81.6) / sqrt(158.90) = -0.84

z2 = (111 - 81.6) / sqrt(158.90) = 1.84

We can then look up the probability of z between -0.84 and 1.84 in a z-table. This value is 0.7652 - 0.3030 = 0.4622.

Therefore, the probability that Y is between 79 and 111 is 0.4622.