SOLUTION: Desertec is a project to supply 15% of Europe's electric power from concentrating solar power systems. When construction begins, large number of mirrors will need to be delivered t

Question 1042675: Desertec is a project to supply 15% of Europe's electric power from concentrating solar power systems. When construction begins, large number of mirrors will need to be delivered to each site to concentrate the suns rays. Suppose you are in charge of a construction site for a 200MW array, requiring 5000 mirrors to be delivered each week over a construction period of 2 years. The delivery company states that the average breakage of mirrors delivered to the site is 1 in every 100 mirrors. (prob of receiving a broken mirror is 0.01) They guarantee that the breakage rate will not exceed this amount.
a) You take a random sample of 10 mirrors from the delivery trucks and find 1 broken mirror. What is the probability of finding 1 or more broken mirrors out of a sample of 10 if the probability of receiving a broken mirror is 0.01?
b) Do you think the delivery company is living up to their guarantee for the situation described in (a)? Give reasons.
c) You take a random sample of 2000 mirrors from the delivery trucks and find 30 broken mirrors. What is the probability of finding 30 or more broken mirrors out of a sample of 2000 if the probability of receiving a broken mirror is 0.01.
d) Do you think the delivery company is living up to their guarantee for the situation described in (c)? Give reasons.

Thanks in advance if you can help me!!!!!!!!
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
probability of 0 broken mirrors is
0.99^10=0.904
The probability of one or more broken mirrors is 0.0996. At a 5% level of significance, this meets the criteria, assuming independence of one mirror from another.
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Expected value is 0.01*2000=20
use alpha=0.05
reject if |z|(can use z because t df=1999) is >1.96
probability of 30 or more mirrors broken
Can use normal probability distribution as well as 1-sample proportion.
First: np=mean=20
sd=sqrt (np(1-p)=sqrt(2000*0.99*0.01)=sqrt(19.8)=4.45
z=(x-mean)/sd=(30-20)/4.45=2.26
p-value=0.0123.
This value is highly significant, the likelihood of this result happening by chance is <2%, and the company is not living up to its guarantee if we set the standard at 5% significance.
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The proportion approach is z=(0.015-0.01)/sqrt{(0.01)(0.99)/2000}
This is 0.005/0.00222=2.245, which is almost the same result.

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