Question 634502: If two percent of electric bulbs manufactured are defective,find the probability that in a sample of 200 bulbs less than 2 bulbs are defective.
[Given: e to the power of -4 =0.0183,e to the power -2=0.114]
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! If two percent of electric bulbs manufactured are defective,find the probability that in a sample of 200 bulbs less than 2 bulbs are defective.
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Binomial Problem with n = 200 and p(defect) = 0.02
P(0<= x <=1) = P(x = 0) + P(x = 1)
= 200C0*0.02^0*0.98^200 + 200C1*0.02*0.98^199 = 1*1*0.01759 + 200*0.02*0.0179
= 0.01759 + 0.0718 = 0.064
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OR, using a TI-84 calculator, you would get:
P(0<= x <=1) = binomcdf(200,0.02,1) = 0.0894
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Cheers,
Stan H.
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