The number of automobiles sold weekly at a certain dealership is a random variable with expected value 16. Give an upper bound to the probability that

(a) next week's sales exceed 18;

(b) next week's sales exceed 25.

(c) Suppose that the variance of the number of automobiles sold weekly is 9. Give a lower bound to the probability that next week's sales are between 10 and 22 inclusively;

(d) give an upper bound to the probability that next week's sales exceed 18.

We use the Markov's inequality to solve for (a) and (b)

P (X > 18) = 16/18 = 8/9 or 0.8888 or 8.88%

P (X > 25) = 16/25 = 0.64 or 64%

For c, we use the z-score with the standard deviation as the square root of the variance

σ = √9 = 3

z = (X - μ) / σ

The limits are 10 and 22

For 10, the z-score is:

z = (10 - 16) / 3 = - 2

For 22

z = (22 - 16) / 3 = 2

We use the z-score table to get the corresponding probability of the two limits and subtract the smaller probability from the bigger probability to get the actual probability. So, from the z-score table:

for z = - 2, P = 0.0228

for z = 2, P = 0.9772

0.9772 - 0.0228 = 0.9544

The probability is 0.9544 or 95.44%

For (d), we do the same thing but we subtract the obtained probability from 1 since the condition is that the sales exceed 18

z = (18 - 16) / 3 = 0.67 which correspond to P = 0.7486

1 - 0.7486 = 0.2514

The probability is 0.2514 or 25.14%