document.write( "Question 1056956:  A manufacturer of window frames knows from past experience that 10 per cent of the production will
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document.write( "have some type of minor defect that will require adjustment. Suppose 30 windows are selected at
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document.write( "random:
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document.write( "a. How many window frames would you expect to have minor defects?
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document.write( "b. What is the probability that more than 3 window frames will need minor adjustments? \n" );
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Algebra.Com's Answer #672034 by Boreal(15235)    You can put this solution on YOUR website! Expected value is 0.10*30=3 window frames x*(p(x)) \n" ); document.write( "probability of 0 is 0.90^30=0.0424 \n" ); document.write( "probability of 1 is 30*0.9^29*0.10=0.1413 \n" ); document.write( "probability of 2 is 435*0.9^28*0.10^2=0.2277 \n" ); document.write( "probability of 3 is 4060*0.9^27*0.1^3=0.2361 \n" ); document.write( "They add to 0.6051 \n" ); document.write( "More than 3 is 1-0.6051=0.3949, using the complement rule. \n" ); document.write( "  |