Find the probability of these 6 example We often use sampling in conjunction wit

Question

Find the probability of these 6 example

We often use sampling in conjunction with deciding the quality of entire shipments. Suppose a shipment of 400 components contains 68 defective and 332 non-defective computer components. From the shipment you take a random sample of 25. When sampling with replacement (so that the p = probability of success does not change), note that a success in this case is selecting a defective part. See instructions below to analyze this situation. NOTE: These are only example questions and the quiz will not be identical to these questions: Exactly two defective components in the sample of 25. Two or more defective components in the sample of 25. Three or less defective components in the sample of 25. Between one and three (inclusive) defective components. . . . What are the mean and variance of this discrete random variable? More than 4 defective components.

Explanation / Answer

BIONOMIAL DISTRIBUTION
pmf of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial
a.
P( X = 2 ) = ( 25 2 ) * ( 0.17^2) * ( 1 - 0.17 )^23
= 0.1193
b.
P( X < 2) = P(X=1) + P(X=0)
= ( 25 1 ) * 0.17^1 * ( 1- 0.17 ) ^24 + ( 25 0 ) * 0.17^0 * ( 1- 0.17 ) ^25
= 0.058
P( X > = 2 ) = 1 - P( X < 2) = 0.942
c.
P( X < = 3) = P(X=3) + P(X=2) + P(X=1) + P(X=0)   
= ( 25 3 ) * 0.17^3 * ( 1- 0.17 ) ^22 + ( 25 2 ) * 0.17^2 * ( 1- 0.17 ) ^23 + ( 25 1 ) * 0.17^1 * ( 1- 0.17 ) ^24 + ( 25 0 ) * 0.17^0 * ( 1- 0.17 ) ^25
= 0.3648
d.
P(1 < = X < = 3 ) = P(X=1) + P(X=2) + P(X=3)
P( X = 1 ) = ( 25 1 ) * ( 0.17^1) * ( 1 - 0.17 )^24
= 0.0486
P( X = 2 ) = ( 25 2 ) * ( 0.17^2) * ( 1 - 0.17 )^23
= 0.1193
P( X = 3 ) = ( 25 3 ) * ( 0.17^3) * ( 1 - 0.17 )^22
= 0.1874
= 0.0486 + 0.1193 + 0.1874
= 0.3553
e.
I.
mean = np
where
n = total number of repetitions experiment is excueted
p = success probability
mean = 25 * 0.17
= 4.25
II.
variance = npq
where
n = total number of repetitions experiment is excueted
p = success probability
q = failure probability
variance = 25 * 0.17 * 0.83
= 3.5275
III.
standard deviation = sqrt( variance ) = sqrt(3.5275)
=1.8782
f.
P( X < 4) = P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 25 3 ) * 0.17^3 * ( 1- 0.17 ) ^22 + ( 25 2 ) * 0.17^2 * ( 1- 0.17 ) ^23 + ( 25 1 ) * 0.17^1 * ( 1- 0.17 ) ^24 + ( 25 0 ) * 0.17^0 * ( 1- 0.17 ) ^25
= 0.3648
P( X > = 4 ) = 1 - P( X < 4) = 0.6352

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