7a) Width of the confidence interval=0.8349-0.6651=0.1698
(2x1.96)((0.75)(1-0.75)/n)^1/2=0.1698
n=100

7b)
(2z) )((0.75)(1-0.75)/100)^1/2=0.1299 Z=1.499955 P(z<1.50)=0.9332
So the new confidence level is 93.32%

8a) Yes, the manufacturer should assume that the average would be 21714 miles driven. It is because the sample mean(21714miles) is usually equal to the population mean. Sample mean is the unbiased estimator of the population mean, so it is appropriate to use it to do further statistical analysis to check every leased car.

8b) Yes, it can choose a confidence level and construct a confidence interval for the average number of miles driven per year. Then the result times $0.3, it will become the confidence interval for the average earning per car. So, the manufacturer can the claim. If $6500 is in between the interval, the claim may be true. 8c) Yes, it is satisfied. There are sample mean(21714miles), standard deviation(2352miles) and the sample size(150).

8d) Yes it does have some concerns. For example, the sample may be chosen from some business which provides services highly relying on car travel, such as delivery services. So, these samples must have a higher averaged miles. As a result, the sample may not representative enough.

8e)Yes, he can. First he constructs a confidence interval for the average number of miles driven per year. Then the result times $0.3, it will become the confidence interval for the average earning per car. The true mean of the earnings may in between this interval. In other words, all numbers in between this interval may be the true mean. So, it is a range of the average amount it can expect to earn per car.

8f) 95% confidence interval for the number of miles driven per year on average: ( 21714-1.96x2352/((150)^1/2) , 21714+1.96x2352/((150)^1/2) )