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Thread: binomial distribution help / confirmation

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    binomial distribution help / confirmation




    Q: A manufacturer produces 1000 computer chips for a mission-critical application. Each chip costs $100 to manufacture and sells for $2000, but has a 1% chance of being defective.

    a) Compute the mean and standard deviation of the number of defective chips.

    b) Assume that 99.9% of defective chips are discovered (and destroyed) before they are sold, while the other 0.1% are sold with defects. The manufacturer suffers an estimated loss of $20,000,000 for each defec- tive chip sold (due to lawsuits, penalties, and bad public relations). What is the manufacturer’s expected profit?

    c) The manufacturer develops a new quality control technology which enables detection of 100% of de- fective chips. How much would the implementation of this technology increase the manufacturer’s expected profit?

    what i think

    a) mean = np =1000*0.01= 10
    standard deviation of the number of defective chips = √npq
    =√1000*0.01*0.99
    =3.146

    b) Total cost of production =1000*100=100000
    Expected number of defectives =10000*0.01 =10
    Expected number of detected =10*99.9/100= 9.99
    Expected number of defectives sold = (10-9.99) =0.01
    Profit =(990*2000)–(1000*100)- 0.01*20000000 =$1680000

    c) The implementation of this technology increase the manufacturer’s expected profit =0.01*20000000 =$200000

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    Re: binomial distribution help / confirmation


    Actually, the company will sell 990 good chips plus .01 defective chips, so the total sales is (990.01*2000)

    I'll let you do the arithmetic

    SO the profit will be a little higher. Since the old expected profit is a little higher, the answer to c will be a little lower.

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