Statistics. Homework help. Please explain.?

The quality control inspector of a production plant will reject a batch of syringes if two or more defective syringes are found in a random sample of eight syringes taken from the batch. Suppose that batch contains 1% defective syringes.





(b) Find µ. What is the expected number of defective syringes the inspector will find?





(c) What is the probability that the batch will be accepted?





(d) Find õ

Statistics. Homework help. Please explain.?
b) this is a binomial distribution and therefore E(X) = n*p = 8*1% = 8% = 0.08





c) P(X%26lt;2) = 1 - P(X=0) - P(X=1)


P(X=0) = 8 nCr 0 *(0.01)^0 * (0.99)^8 = 0.99^8 = 0.9227


P(X=1) = 8 nCr 1 *(0.01)^1 * (0.99)^7 = 8*0.01*(0.99)^7 = 0.07456


1 - P(X=0) - P(X=1) = 1 - 0.9227 - 0.07456 = 0.00269





d) I think that's supposed to be a sigma and therefore the standard deviation: for binomial distribution variance = n*p*(1-p) = 8*0.01*0.99 = 0.0792 Therefore the standard deviation is sqrt(0.0792) = 0.2814
Reply:Let X be the number of defective syringes. X has the binomial distribution with n = 8 trials and success probability p = 0.01





In general, if X has the binomial distribution with n trials and a success probability of p then


P[X = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[X = x] = 0 for any other value of x.





The probability mass function is derived by looking at the number of combination of x objects chosen from n objects and then a total of x success and n - x failures.


Or, in other words, the binomial is the sum of n independent and identically distributed Bernoulli trials.





X ~ Binomial( n , p )





the mean of the binomial distribution is n * p = 0.08


the variance of the binomial distribution is n * p * (1 - p) = 0.0792


the standard deviation is the square root of the variance = √ ( n * p * (1 - p)) = 0.2814249





The Probability Mass Function, PDF,


f(X) = P(X = x) is:





P( X = 0 ) = 0.9227447


P( X = 1 ) = 0.07456523


P( X = 2 ) = 0.002636144


P( X = 3 ) = 5.325544e-05


P( X = 4 ) = 6.724172e-07


P( X = 5 ) = 5.433674e-09


P( X = 6 ) = 2.74428e-11


P( X = 7 ) = 7.92e-14


P( X = 8 ) = 1e-16





The Cumulative Distribution Function, CDF,


F(X) = P(X ≤ x) is:





x


∑ P(X = t) =


t = 0





P( X ≤ 0 ) = 0.9227447


P( X ≤ 1 ) = 0.99731


P( X ≤ 2 ) = 0.999946


P( X ≤ 3 ) = 0.9999993


P( X ≤ 4 ) ≈ 1


P( X ≤ 5 ) ≈ 1


P( X ≤ 6 ) ≈ 1


P( X ≤ 7 ) ≈ 1


P( X ≤ 8 ) = 1








1 - F(X) is:





n


∑ P(X = t) =


t = x





P( X ≥ 0 ) = 1


P( X ≥ 1 ) = 0.0772553


P( X ≥ 2 ) = 0.002690078


P( X ≥ 3 ) = 5.393332e-05


P( X ≥ 4 ) = 6.778784e-07


P( X ≥ 5 ) = 5.461196e-09


P( X ≥ 6 ) = 2.752210e-11


P( X ≥ 7 ) = 7.926992e-14


P( X ≥ 8 ) = 1.110223e-16





The probability that the batch will be rejected is:





P( X ≥ 2 ) = 0.002690078

sending flowers