STATS help please show work and equations/ Formulas Problem A Questions 1-3 An i

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STATS help please show work and equations/ Formulas

Problem A Questions 1-3 An investor's portfolio consists of two stocks: A and B. The following represeats each stock's rate of return (in %) for a sample of five periods. Period1 Period2 Period3 Period4 Periods Stock A Stock B -4.8 8.9 9.33.66.08. 1.2 1. Find the 1QR for Stock B's rate of return. 2. Find the covariance 3. Find the corrvlation. Below are the annual household incomes (in $1000s) of 15 randomly selected U.S. households 46.2, 58.1, 59.7, 43.6, 47.2, 50.3, 41.5, 64.6, 62.4, 57.8, 55.4, 60.1, 49s 40.9, 6 (SEE EXCEL OUTPUT, APPENDIX 1) 1. Suppose the average age of the head of the household for this sample was 42.57 years with a standard deviation of 7.93 years. When comparing the most? What is the value of this statistic? income and age, which varied 2. What is the shape of the income distribution? Rule predicts at percent of the income data should lie between 3. The 528,814.30 and 577,959.10. Problem C. Questions 1-3 An inspector randomly selects three parts from a lot that costains 2 defective and 4 nondefective 1. What is the probabálity that the inspector finds exactly one defective part? 2. What is the probability that the inspector finds exactly one nondefective part 3. What is the probability that the inspector finds exactly two defective parts? parts 17

Explanation / Answer

Problem C:

Number of defective parts: 2

Number of non defective parts: 4

Total number of parts: 6

Number of ways of selecting 3 parts out of 6 is

C(6,3) = 20

(a)

Number of ways fo selecting 1 defective part and 2 non defective parts is

C(2,1)C(4,2) = 2 * 6 = 12

Therefore the probability of selecting 1 defective part is

P(exactly 1 defective part) = 12/20 = 0.6

(b)

Number of ways fo selecting 2 defective parts and 1 non defective parts is

C(2,2)C(4,1) = 4

Therefore the probability of selecting 1 nondefective part is

P(exactly 1 nondefective part) = 4/20 = 0.2

(c)

Number of ways fo selecting 2 defective parts and 1 non defective parts is

C(2,2)C(4,1) = 4

Therefore the probability of selecting 2 defective part is

P(exactly 2 defective parts) = 4/20 = 0.2

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