# SOLUTION: I have a few questions that I am having a very difficult time with. 1.) A 28-year-old man pays \$ 175 for a one-year life insurance policy with coverage of \$140,000. If the pro

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: I have a few questions that I am having a very difficult time with. 1.) A 28-year-old man pays \$ 175 for a one-year life insurance policy with coverage of \$140,000. If the pro      Log On

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 Click here to see ALL problems on Probability-and-statistics Question 191844: I have a few questions that I am having a very difficult time with. 1.) A 28-year-old man pays \$ 175 for a one-year life insurance policy with coverage of \$140,000. If the probability that he will live through the year is 0.9993, what is the expected value for the insurance policy? 2.)Two dice are rolled. Find the odds that the score on the dice is either 10 or at most 5. 3.)What are the odds in favor of getting at most two heads in three successive flips of a coin? 4.)Assume that the weight loss for the first two months of a diet program has a uniform distrbiution over the interval 6 to 12 pounds. Find the probability that a person on this diet loses between 9 and 12 pounds in the first two months. Answer by stanbon(57361)   (Show Source): You can put this solution on YOUR website!1.) A 28-year-old man pays \$ 175 for a one-year life insurance policy with coverage of \$140,000. If the probability that he will live through the year is 0.9993, what is the expected value for the insurance policy? ------- Ramdom Variable X = -175 , 140,000 Corresponding Probabilities: 0.9993, 0.0007 ------------------------ Expected Value: 0.9993(-175) + 0.0007(140,000) = -\$76.88 ===================================================================== 2.)Two dice are rolled. Find the odds that the score on the dice is either 10 or at most 5. Sketch the sample space (a 6x6 square of dice sums) and you will see that P(10) = 3/36 and P(1= P(10 or <=5) = 13/36 Therefor the odds are [13/36]/[23/36] = 13/23 ====================================================================== 3.)What are the odds in favor of getting at most two heads in three successive flips of a coin? P(0<=x<=2) = 1 - P3 heads) = 1-(1/8) = 7/8 Therefore the odds in favor = (7/8)/(1/8) = 7:1 ====================================================================== 4.)Assume that the weight loss for the first two months of a diet program has a uniform distrbiution over the interval 6 to 12 pounds. Find the probability that a person on this diet loses between 9 and 12 pounds in the first two months. Draw the picture: The mean value is 9. The right tail is 12. Half the population is above the interval 9 to 12. The Probibility is 50% = 0.5000 =================================== Cheers, Stan H.