For a certain individual, at breakfast the calorie intake is arandom variable wi
ID: 2918604 • Letter: F
Question
For a certain individual, at breakfast the calorie intake is arandom variable with an expected value of 500 and standarddeviation of 50. The calorie intake at lunch is random and has anexpected value of 900 and standard deviation of 100. The calorieintake at dinner is also a random variable and has an expectedvalue of 2000 and a standard deviation of 180.Assuming that the intakes at the different meals areindependent of each other, what will be the probability that theaverage calorie intake per day over the next year (365 days) is atmost 3500?
(Hint: Have Xi, Yi, Zi denote the three calorie intakes on dayi. The total intake is given by sum(Xi + Yi + Zi)
Assuming that the intakes at the different meals areindependent of each other, what will be the probability that theaverage calorie intake per day over the next year (365 days) is atmost 3500?
(Hint: Have Xi, Yi, Zi denote the three calorie intakes on dayi. The total intake is given by sum(Xi + Yi + Zi)
Explanation / Answer
Let Ci = (Xi + Yi +Zi )denote calorie intake on a day i.
P( average of Ci over 350 days is < 3500)
= P( average of (Xi + Yi +Zi) < 3500)
E(Xi + Yi +Zi)= E(Xi)+E(Yi)+E(Zi) (since X,Y,Z are given asindependent) = 500+900+2000 = 3400.
Var (Xi + Yi +Zi) = Var (Ei) +Var(Yi) +Var(Zi) (again becauseX,Y,Z are independent) =502+1002+1802 = 44900.
Since 350 is a large sample average of (Xi + Yi +Zi)follow as normal distribution with mean = 3400 and stdev =sqrt(44900)/sqrt(350) = 11.32
there fore P(average of (Xi + Yi +Zi)) < 3500)
= P(z< (3500- 3400)/11.32 )
= P(z < 8.82) = 99.99999%
Hope this helps. Feel free to ask for anyclarifications.