Assume that you assign the following subjective probabilities for your final grade in your econometrics course (the standard GPA scale of 4-A to 0-F applies): Grade A The expected value is: OA. 2.78. OB. 3.5. OC. 3.0. OD. 3.25. (BUDE с F Probability 0.20 0.50 0.20 0.08 0.02
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- A department store manager has monitored the number of complaints received per week about poor service. The probabilities for numbers of complaints in a week, established by this review, are shown in the table. Number of complaints Probability 0 0.16 1 0.26 2 0.35 week? 3 0.10 4 0.08 5 0.05 What is the standard deviation of complaints received per Please round your answer to the nearest hundredth. Note that the correct answer will be evaluated based on the full-precision result you would obtain using Excel.The average student loan debt for college graduates is $25,750. Suppose that that distribution is normal and that the standard deviation is $10,350. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. Find the probability that the college graduate has between $22,700 and $30,000 in student loan debt. (------) The middle 20% of college graduates' loan debt lies between what two numbers? Low: $(______) High: $(_____)The average student loan debt for college graduates is $25,600. Suppose that that distribution is normal and that the standard deviation is $13,450. Let X = the student loan debt of a randornly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. a. What is the distribution of X? X - NO b Find the probability that the college graduate has between $10,750 and $15,950 in student loan debt. c. The middle 20% of college graduates' loan debt lies between what two numbers? Low: $ High: $ Submit Question M Co
- The following data are the monthly salaries and the grade point averages for students who obtained a bachelor's degree in mathematics. GPA: 2.6, 3.4, 3.6, 3.2, 3.5, and 2.9; Monthly Salary: 3300, 3600, 4000, 3500, 3900, and 3600. Develop a 95% prediction interval for the expected value of y when x 3.6. Explain.In the first quarter of 2017, the average mortgage for first time buyers was R910,000. Assuming a normal distribution and a standard deviation of R50,000. What proportion of mortgages were between R890,000 and R990,000 (rounded off to three decimals)? A. 0.601 B. 0.345 C. 0.945 D. 1.600Suppose the returns on an asset are normally distributed. The historical average annual return for the asset was 6.6 percent and the standard deviation was 16.5 percent. a. What is the probability that your return on this asset will be less than -8.5 percent in a given year? Use the NORMDIST function in Excel® to answer this question. Note: Do not round intermediate calculations and enter your answer as a percent rounded to 2 decimal places, e.g., 32.16. b. What range of returns would you expect to see 95 percent of the time? Note: Enter your answers for the range from lowest to highest. A negative answer should be indicated by a minus sign. Do not round intermediate calculations and enter your answers as a percent rounded to 2 decimal places, e.g., 32.16. c. What range of returns would you expect to see 99 percent of the time? Note: Enter your answers for the range from lowest to highest. A negative answer should be indicated by a minus sign. Do not round intermediate calculations and…
- 0. 0.02 1 0.10 0.36 3 0.52 M:2.38 0= 0.75 7. For the probability distribution X~B(39,0.72), find the mean and the standard deviation. P.5 Expected Value 8. You play a game by drawing a card from a standard deck of cards. You must pay $3 to pla If you draw an Ace, you win $25. If you draw a King, you win $10. If you draw any other card, you win nothing. Find the expected value. Is this a good game for you to play? (Not« There are 4 Aces and 4 Kings in a standard deck of 52 cards).Suppose a life insurance company sells a $170,000 1-year term life insurance policy to a 20-year-old female for $170. According to the National Vital Statistics Report, 58(21), the probability that the female survives the year is 0.999544. The expected value of this policy to the insurance company is $92.48. What is the standard deviation of the value of the life insurance policy? Why is the value so high?The average student loan debt for college graduates is $25,750. Suppose that that distribution is normal and that the standard deviation is $13,900. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. a. What is the distribution of X? X ~ N( b Find the probability that the college graduate has between $20,750 and $39,650 in student loan debt. c. The middle 30% of college graduates' loan debt lies between what two numbers? Low: $ High: $
- The average student loan debt for college graduates is $25,200. Suppose that that distribution is normal and that the standard deviation is $13,300. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar. a. What is the distribution of X? X - N b Find the probability that the college graduate has between $22,700 and $29,650 in student loan debt. c. The middle 30% of college graduates' loan debt lies between what two numbers? Low: $ High: STyson Corp. stock has returns of 3%, 18%, -24%, and 28% for the past four years. Based on this information, what is the 99% probability range for any one given year? Select one: Oa 24.5 to 34.3% O b.-61.7 to 74.2% OC.-8.4 to 11.7% Od.-16.4 to 28.9% e. 39.0 to 51.5%Suppose a life insurance company sells a $170,000 1-year term life insurance policy to a 20-year-old female for $300. According to the National Vital Statistics Report, 58(21), the probability that the female survives the year is 0.999544. The expected value of this policy to the insurance company is$222.48. What is the standard deviation of the value of the life insurance policy? Why is the value so high? Part 1 The standard deviation of the value of the life insurance is $