Student Name: ba ( Student ID: Question 4: 125 Marks) Assume that the weights of

Question

Student Name: ba ( Student ID: Question 4: 125 Marks) Assume that the weights of adult individuals are independent and Uniformly distributed between 60 kg to 100 & Suppose that 35people squeeze into an elevator that is designed to hold 2.900 Kg. ,,- i) 8pts) what is the average and standard deviation of the total weight of the 35 people? And what is the probability density function A") of their total weight (justify your answer)? ECK): /2 ,f(w) 2H0 -a 5pts) what is the probability that the load (total regt) exceeds the design limit of the elevator? 2900 ii) (8pts) Assume that electricity costs 0.1 QAR per kg for the first 15 persons and it is 0.2 QAR per kg for any extra person getting into the elevator after the first 15 persons. Find the mean of the total electricity cost and its variance for the 35 people who squeezed inside the elevator. x)(y-15)-2.52 /2 minimum number of people that we must remove from the elevator out of the 35 people? [Hint: Try expressing the mean and the variance as a function of the mumber persons.] iv) (4pts) If we want the probability of exceeding the elevator specification to be less than 0.05, what is the

Explanation / Answer

X - unif (60,100)

E(X) = 80 , Var(X) = 40^2 /12

sd(X) = 20/sqrt(3)

S = X1+X2+...X35

E(S) = 35*E(X) = 35*80 = 2800

sd(S) = sqrt(n * Var(X)) = sqrt(35 )*20/sqrt(3) = 68.3130

b)

P(S > 2900)

since n >30, we can use central limit theorem

Z = (S - 2800)/68.3130

P(S > 2900)

=P(Z> (2900 - 2800)/68.3130)

=P(Z> 1.46385)

= 0.0716

c)

0.1/kg for first 15 person

0.2/keg for next 20 person

Y = 0.1 (X1 + X2+..X15) + 0.2*(X16 +...X35)

E(Y) = 0.1* 15*E(X) + 0.2*20* E(X)

=(1.5 + 4 )*80

=440

Var(Y) = 0.1* 15* var(X) + 0.2*20* var(X)

= (1.5 +4)* 40^2 /12

=733.33333

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