Question

2. (-/1 Points) DETAILS DEVORESTAT9 5.E.079.NVA MY NOTES PRACTICE ANOTHER Suppose that for a certain individual, calorie intake at breakfast is a random variable with expected value 600 and standard deviation 40, calorie intake at lunch is random with expected value 900 and standard deviation 100, and calorie intake at dinner is a random variable with expected value 2100 and standard deviation 180. Assuming that intakes at different meals are independent of one another, what is the probability that average calorie intake per day over the next (365-day) year is at most 3700? (Hint: Let X, Y, and z, denote the three calorie intakes on day i. Then total intake is given by X, + Y, + Z;). Round your answer to four decimal places.) You may need to use the appropriate table in the Appendix of Tables to answer this question. Need Help? Talk to a Tutor

Answer 1

Solution Given Xi and 2) denole the three calorie Inlabel On dag i So the total Calorie (say reperesented by (x)) is given by х Xit Yitzi = E(X) +E() i) + Elzs G(Xi) = 6oo SD (Xi) = {(4) - 900 E (4) Now yo SD(Yi ) =100 G(Zi) = 2100 SDiZij = 180 M = E(X) = boot 900 + 2100 = 3600 -V(X) = V(Xi) + V1 Yi) + V121 02= + (100)2 + (180) 44000 प५००० 14019 aa J09769

PC Averase intake per day se ad most 370 ) PC x < 3700) X-M pl U 3700-3600 9.9.76 Plz P z L 0.47673) 0.68399 = 0 68 322 0:6832 0.0 68 32
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