Question

You want to use a set of wires to suspend a weight for the ceiling of your workshop. After many tests, you reach the conclusion that each wire used has a breaking point given by a random variable X with E(X) 10kg and ?x--2kg. You must suspend a weight of 500kg so you decide to use a large number of wires to suspend it. Suppose that the breaking point of a set of wires used together is the sum of the breaking points of each of them and that each wire is independent from the others. Compute the probability the the wires will not break when you use 50, 54 or 58 of them. (Hint: since 50 is large, you can use the CLT.)

Answer 1

for 50,
mean = 50*10 = 500
sd =2*sqrt(50) = 14.1421

P(X < 500)
= P(z < (500 - 500)/14.14)
= P(z < 0)
= 0.5

Hence 0.5 is the probability that the wires will not break when we use 50 of them.

for 54,
mean = 54*10 = 540
sd =2*sqrt(54) = 14.6969

P(X > 500)
= P(z > (500 - 540)/14.6969)
= P(z > -2.722)
= 0.9968

Hence 0.9968 is the probability that the wires will not break when we use 54 of them.

for 58,
mean = 58*10 = 580
sd =2*sqrt(58) = 14.6969

P(X > 500)
= P(z > (500 - 580)/15.2315)
= P(z> -5.2523)
= 1

Hence 1.0000 is the probability that the wires will not break when we use 58 of them.