(C Suppose that a day\'s production of 850 manufactured parts are selected at ra
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Question
(C Suppose that a day's production of 850 manufactured parts are selected at random, Sampling Without Replocement contains 50 parts that do not conform to customer requirements. Two parts Example 3-8 without replacement, from the batch. Let the random variable X equal the number of nonconforming parts in the sam- ple. What is the cumulative distribution function of X? The question can be answered by first finding the probability mass function of X. 800 799 0.886 PX)850 849 P(X = 1) 2.850 849 P(X = 2)= 850, 849 800.50=0.111 5049 = 0.003 Therefore F(0) = P(X F(I) P(X F(2) = P(X 0) = 0.886 1 ) = 0.886 + 0.1 1 1 = 0.997 2) = 1 and 0 0.886 0sxExplanation / Answer
1)probability that first part is conforming =800/850 ( as there are 800 conforming part out of 850)
if first part selected is conforming ; then there left 799 conforming part out of remaining 849 parts
P(X=0) =P(both parts are conforming)=P(first part is conforming and second part is conforming)
=(800/850)*(799/849)=0.886
2)
probability that first part is non conforming =50/850 ( as there are 50 non conforming part out of 850)
if first part selected is non conforming ; then there left 800 conforming part out of remaining 849 parts
P(X=1) =P(one part conforming and one part is non conforming)
=P(first part conforming and second non conforming+first part non conforming and second conforming)
=(800/850)*(50/849)+(50/850)*(800/849) =0.111
P(X=2) =P(both part are non conforming) =(50/850)*(49/849)=0.003
please revert if further clarification is required
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