binomial distribution help / confirmation

#1
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
 
#2
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.