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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

Anext week's sales exceed 18;

(b )next week's sales exceed 25.

CSuppose 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;
Dgive an upper bound to the probability that next week's sales exceed 18.

Verified Answer :

MissPhiladelphia

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%

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