2 Consider the following data on new customers for AJ Auto I

2. Consider the following data on new customers for AJ Auto Insurance, specifically the information of the risk level of the customer and the number of tickets they have had in the last year.

0

1

2 or more

Total

Low Risk

56

22

8

86

Medium Risk

18

40

12

70

High Risk

11

13

20

44

Total

85

75

40

200

If you choose a customer at random, then find the probability that the customer is
a. low risk.
b. low risk and had two or more tickets in the last year.
c. medium risk, given that the customer had zero tickets in the last year.

3. DCW Chemical is planning to implement an acceptance sampling plan for raw materials. A random sample of 22 batches from a large shipment (having a large number of batches) is selected. If two or more of the 22 batches fail to meet specifications, then the entire shipment is returned. Otherwise, the shipment is accepted.

In a sample of 22 batches from a population that is 1% defective (1% of the batches fail to meet specifications), find the probability that
a. two or more batches fail to meet specifications.
b. exactly two batches fail to meet specifications.
c. fewer than two batches fail to meet specifications.  

5 A survey is taken of U.S. companies that do business with firms in India. One of the questions in the survey was: “Approximately how many years has your company been trading with firms in India?” A random sample of 44 responses to this question yields the following results.

Sample Size = 44
Sample Mean = 10.46 years
Sample Standard Deviation = 7.7 years

a. Compute the 95% confidence interval for the population mean number of years of trading with firms in India.
b. Interpret this interval.
c. How many firms should be sampled if we wish to generate a 95% confidence interval for the mean number of years of trading with firms in India that is accurate to within 1 year?

Solution

a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=10.46
Standard deviation( sd )=7.7
Sample Size(n)=44
Confidence Interval = [ 10.46 ± Z a/2 ( 7.7/ Sqrt ( 44) ) ]
= [ 10.46 - 1.96 * (1.16) , 10.46 + 1.96 * (1.16) ]
= [ 8.18,12.74 ]
b)
We are 95% confident that the population is mean is lies
in this interval
c)
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 7.7
ME =1
n = ( 1.96*7.7/1) ^2
= (15.09/1 ) ^2
= 227.77 ~ 228