SOLUTION: From past experience, a company found that in cartons of DVDs, 90% contain no defective DVDs, 5% contain one defective DVD, 3% contain two defective DVDs, and 2% contain three defe

Question 1051786: From past experience, a company found that in cartons of DVDs, 90% contain no defective DVDs, 5% contain one defective DVD, 3% contain two defective DVDs, and 2% contain three defective DVDs. Find the mean, variance, and standard deviation for the number of defective DVDs. Please explain all steps.
Thank you!
Found 2 solutions by stanbon, Boreal:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
From past experience, a company found that in cartons of DVDs, 90% contain no defective DVDs, 5% contain one defective DVD, 3% contain two defective DVDs, and 2% contain three defective DVDs. Find the mean, variance, and standard deviation for the number of defective DVDs. Please explain all steps.
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Percent of defective = 100%-90% = 10%
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mean = n*p = 0.1*n
variance = n*p*q = 0.1n*0.9 = 0.09n
std = sqrt(npq) = sqrt(0.09n) = 0.3*sqrt(n)
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Cheers,
Stan H.
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Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
x====0====1====2====3
p(x)0.9=0.05 0.03==0.02
E(x), the mean is x*p(x)=0+0.05+0.06+0.06=0.17 defect per carton.
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variance is the average (here, a weighted average) of the squared deviations of the mean. This is [x-E(x)]^2*p(x)
0.17^2*0.90=0.0260
0.83^2*0.05=0.0344
1.83^2*0.03=0.1098
2.83^2*0.02=0.1602
Sum is 0.3304 defective^2/box^2
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Standard deviation is sqrt(V(x))=0.5748 defective/box
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