An upper bound on the probability that at least 175 products are made before a defect is discovered is 0.57 (Option C)
According to Markov's inequality, the chance that a positive real number is larger than or equal to a given positive random variable X is either less than or equal to the expected value of X divided by a.
Let X stand for the random variable that represents the first flaw found; X has a geometric distribution with p=0.01.
Then we'll have: E(X) = (1 - p) / p = (1 - 0.01) / 0.01 = 99
Markov's inequality has been used to create:
P(X ≥ 175) = E(X) / 175 = 99 / 175 = 0.5657 ≅ 0.57
Therefore, the highest limit on the likelihood that at least 175 goods be produced before a flaw is found is:
P(X ≥ 175) = 0.57
Therefore, option C is the correct choice.
To know more about Markov's inequality, refer to this link:
brainly.com/question/28902943
#SPJ4
<u>COMPLETE QUESTION:</u>
Suppose a production line stops maintenance whenever a defective product is produced. If there is a 1% chance of a defective product, what is an upper bound on the probability that at least 175 products are made before a defect is discovered? (Hint: Find an upper bound, NOT the actual probability.)
a. 0.34
b. 0.01
c. 0.57
d. 0.17
