Arandom sample of 50 units is drawn from a production process every half hour. the true fraction of nonconforming products manufactured is 0.03. what is probability that the estimated probability of a nonconforming product will be less than or equal to 0.06 if the fraction of nonconforming really is 0.03

solution:

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

find the probability that (p∧ ≤ 0.06) , substitute the values of sample units (n) , and the probability of nonconformities (p) in the probability mass function of binomial distribution.

Consider x   to be the number of non-conformities. It follows a binomial distribution with n   being 50 and p  being 0.03. That is,

binomial (50,0.02)

Also, the estimate of the true probability is,

p∧  = x/50

The probability mass function for binomial distribution is,

Where,

X=0,1,2,3,…..; q=1-p

The calculation is obtained as

P(p^ ≤ 0.06) = p(x/20 ≤ 0.06)

         = 50cx ₓ (0.03)x ₓ (1-0.03)50-x  

=    (50c0 ₓ (0.03)0 ₓ (1-0.03)50-0 + 50c1(0.03)1 ₓ (1-0.03)50-1 + 50c2 ₓ (0.03)2 ₓ (1-0.03)50-2 +50c3 ₓ      (0.03)3 ₓ (1- 0.03)50-3 )

=(   ₓ (0.03)0 ₓ (1-0.03)50-0 +  ₓ (0.03)1 ₓ (1-0.03)50-1 +   ₓ (0.03)2 ₓ (1-0.03)50-2   ₓ (0.03)3 ₓ (1-0.03)50-3 )

answer: idk

step-by-step explanation:

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