Question

Please state general strategy needed, state formulas.

A store's revenue in a day can be described by a random variable, with a mean of USD 100,000 and a standard deviation of USD 50,000

An upper limit on the probability that the store has an income of more than USD 250,000 on one day, is?

The answer is not probability = 0.0013, it is 1/9, help me find the required methodology to reach this answer.

Answer 1

here from Tchebychev Inequality:

P(X-$\mu$ >k) <=$\sigma$² /k²  

from above $\mu$ =100000 and $\sigma$ =50000  

therefore P(X>250,000) =P(X-100000 >150000) <=50000²/150000²

or P(X>250,000) <=1/9