2. The probability that the insured will incur a loss is 0.05. If there is a loss, = probability of a loss of amount N is K 'N for N = 1, .., 8 and K a constant. These are the only possible loss amounts d no more than one loss can occur Determine the net premium for this policy

4

Suppose X~N(177,17) and Z~N(0,1). If z = 0.75, what is it on an X scale? Express your answer to two decimal places (ex: 123.45).

please solution about C D and E

H H T T T H T H H T TH T H T H TH TH TH T The probability of obtaining two heads and two tails (Type an integer or a fraction.)

15. In January 2013, the dear population in a certain area was counted at 3,000. Since then, the population has decreased by about 12% each year. In which year will the population be expected to drop below 500? A. 2025 ACEA B. 2026 C. 2027 D. 2028 O O O A B E: All of the Above F: None of the Above

state s probability P(s) realization X(s) realization Y(s) 1 .1 10 20 2 .2 20 30 3 .2 10 10 4 .3 10 30 5 .1 20 10 6 .1 30 20 What is E(Y|X = 20)? a) 65/3 b) 70/3 c) 25 d) 80/3

9(D)

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x, is found to be 114, and the sample standard deviation, s, is found to be 10. (a) Construct a 96% confidence interval about µ if the sample size, n, is 21. (b) Construct a 96% confidence interval about µ if the sample size, n, is 12. μ (c) Construct an 80% confidence interval about µ if the sample size, n, is 21. (d) Could we have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed? Click the icon to view the table of areas under the t-distribution. (a) Construct a 96% confidence interval about µ if the sample size, n, is 21. Lower bound: Upper bound: (Use ascending order. Round to one decimal place as needed.)

4. 65th Pg. 59, 1. 7. 19. * Find the of the distribution having pot F(x) = 1x1/4₂ Percentile where -a, < x < 2, f(x) = 0, elsewhere.