1. The monthly sales of mufflers in the Richmond, VA area follow the normal dist
ID: 2958522 • Letter: 1
Question
1. The monthly sales of mufflers in the Richmond, VA area follow the normal distribution with a mean of 1200 and a standard deviation of 225. The manufacturer would like to establish inventory levels such that there is only a 5% chance of running out of stock. Where should the manufacturer set the inventory levels?2.Research on new juvenile delinquents revealed that 38% of them committed another crime.
a. What is the probability that of the last 100 new juvenile delinquents put on probation, 30 or more will commit another crime?
b. What is the probability that 40 or fewer of the delinquents will commit another crime?
c. What is the probability that between 30 and 40 of the delinquents will commit another crime?
3. An Air Force study indicates that the probability of a disaster such as the January 28, 1986 explosion of the space shuttle Challenger was 1 in 35. The Challenger flight was the 25th mission.
a. How many disasters would you expect int he first 25 flights?
b. Use the normal approximation to estimate the probability of at least one disaster in 25 missions.
Explanation / Answer
1.X= monthly sales
Let the levels 1200-k and 1200+k so we have to find k where
P(1200-k<1200+k)=0.95
P(-k/225
P(Zk/225 = 1.96
k=1.96*225=441
Levels are 1200-441=759 and 1200+441=1641
Answer: 759 and 1541
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2.Let X= number of delinquents that commit another crime in the last 100 delinquents
X has a binomial distribution with n=100 and p=0.38
We have to find
a) P(X30)
b) P(X40)
c) P(30<40)
We can use normal approximation:
Y has a normal distribution with mean = np = 38
And sd = [np(1-p)] = [100*0.38*0.62]=4.854
a) P(X30) P(Y>29,5)=P(Z>(29,5-38)/4.854)=P(Z>-1,751)=0.96
b) P(X40) P(Y<40,5)=P(Z<(40,5-38)/4.854)=P(Z<0.515)=0.697
c) P(30<40) P(30.5<39.5)= P((30.5-38)/4.854<(39.5—38)/4.854)
=P(-1.545<0.309)=0.56
Answer:
a) 0.96
b) 0.697
c) 0.56
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3.X= number of disasters in 25 missions
X has a binomial distribution with n=25 and p=1/35
a) E(X)=np= 25/35=5/7
b) P(X1) = P(Y>0.5) where Y has a normal distribution with mean = 5/7
And sd = [np(1-p)] = [25*(5/7)*(2/7)]=5.102
P(Y>0.5)=P(Z>(0.5-5/7)/5.102)=P(Z>-0.042)=0.517
Answer:
a) 5/7
b) 0.517
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