Respuesta :
Answer:
The number of defective bulbs in the shipment should be around 15, give or take 4.
Step-by-step explanation:
For each bulb, there are only two possible outcomes. Either it is defective, or it is not. The probability of a bulb being defective is independent of any other bulb. This means that the binomial probability distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x successes on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Suppose the probability at a light bulb factory of a bulb being defective is 0.11
This means that [tex]p = 0.11[/tex]
Shipment of 133 bulbs:
This means that [tex]n = 133[/tex]
Mean and standard deviation:
[tex]E(X) = np = 133*0.11 = 14.63[/tex]
[tex]\sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{133*0.11*0.89} = 3.61[/tex]
Rounding to the nearest integers:
The number of defective bulbs in the shipment should be around 15, give or take 4.
