Using Central Limit Theorem,
the smallest possible value of sample size , n is 452...
let Xi = 1 if item is defective or
Xi = 0 if item is not defective
Now, Xn- hat = 1/n ∑Xᵢ= sample proportion for defective items
we have to calculate smallest value of n we require P( Xn- hat <0.13 ) > 0.99
we have,. E(Xᵢ) = μ = 0.1 and variance (Xᵢ)
= (.1 ) (1 - 0.1) =( 0.1) (0.9) = σ²
Since, Xi are Bernoulli random variables
According to limit theorem ,
Z = (Xₙ- hat - u) √n/ sd ~ N(1;0) approximately when n is large
Therefore,
P ( Xₙ -hat < 0.13)
=> P (( Xₙ-hat - u)√n / sd < (0.13 - 0.1 )√n/ √0.09 )
=> P( Z < 0.03 √n/0.03 ) where Z~N(1;0)
=> P( Z< √n/10)
i.e f( √n/10 > f(2.32 ) = 0.99
=> n/ 100 > (2.32)²
=> n = (2.32)²× 100
=> n > 541.02 ~ 542
Thus, smallest possible value is 542..
To learn more about Center limit theorem, refer:
https://brainly.com/question/14099217
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