This question comes from the \"Discrete Random Variables and Their Probability D

Question

This question comes from the "Discrete Random Variables and Their Probability Distributions" chapter. If a production process is running properly, 10% of the bearings fail to meet the required specifications. Sometimes problems develop with the production process that causes the rejection rate to exceed 10%. To guard against this higher rejection rate, samples of 15 bearings are taken periodically and carefully inspected. If more than 2 bearings in a sample of 15 fail to meet the required specifications, production is suspended for necessary adjustments.

If the true rate of rejection is 10% (that is, the production process is working properly), what is the probability that the production will be suspended based on a sample of 15 bearings?

What assumptions did you make?

Explanation / Answer

Note that P(more than x) = 1 - P(at most x).          
          
Using a cumulative binomial distribution table or technology, matching          
          
n = number of trials =    15      
p = the probability of a success =    0.1      
x = our critical value of successes =    2      
          
Then the cumulative probability of P(at most x) from a table/technology is          
          
P(at most   2   ) =    0.815938931
          
Thus, the probability of at least   3   successes is  
          
P(more than   2   ) =    0.184061069 [ANSWER]

**********************

Here, I assumed that the bearings are independent from each other, and the probability of defect is constant at 0.10.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.