Question

4. A lift has an occupancy warning of no more than 25 people and of total weight no more than 1950kg. For a population of users, suppose weights are normally distributed with mean 75kg and standard deviation 10kg.

(a) What is the probability that the total weight of a random sample of 25 people from the population exceeds 1950kg?

(b) Calculate the probability that a random sample of 24 people sets the alarm off.

(c) Suppose people carry things with them and the weight Y of all goods for a person with weight X in kg is R(1, 7).

i. Let U = X + Y , calculate mean and variance of U.

ii. What is an approximate probability that a random sample of 25 people in the lift exceeds 1950kg?

Answer 1

Answer:

giventhat

a)
S = X1+ X2+..X25
E(S) = 25 E(X) = 25 * 75 = 1875
sd(S) = sqrt(n) * sd(X) = 10*sqrt(25) = 50

P(S > 1950)
= P(Z > (1950-1875)/50)
= P(Z > 1.5 )
= 0.0668

b)
T = X1+ ...X24
P(T > 1950)
= P(Z > (1950 - 24 * 75)/(10*sqrt(24))
= P(Z > 3.06186 )
= 0.0011

c)
i) X = N(75,10^2)
Y = N(1,7)
X+Y = N(75 +1 , 10^2 +7)
= N(76,107)

S = X1+ ..X25
P(S > 1950)
= P(Z >(1950 - 25*76)/(sqrt(107 * 25))
= P(Z > 0.96673648904 )
= 0.1668