Question

Please just do part i.

Vol<sWasen A shipping company has a large fleet of Peterbilt, T trucks. The distribution of weights in lb of these trucks from Peterbilt, Volkswagen and Ford are N(580, 225), N (100,64) and N(120, 81), respectively, and are independent. i. If one Peterbilt truck, three Volkswagen trucks and a Ford truck are chosen in dependently and at random, what is the probability that twice the weight of the Ford truck and the sum of the weights of the three Volkswagen trucks is greater than the weight of the Peterbilt truck? ii. Find the probability that one-third the weight of a Peterbilt truck or twice the weight of a Ford truck or five-thirds the weight of a Volkswagen truck is less than 200 ib iii. A random sample of 12 Peterbilt trucks is to be chosern a. Find the probability that the sample mean will be greater than 573.505 and less than 578.701 b. With S2 t he sample variance, find constants a and b so that P(a < S2 s b) 0.90.

Answer 1

Let P be the random variable denoting the weights of Peterbilt trucks

Let V be the random variable denoting the weights of Volkwagen trucks

Let F be the random variable denoting the weights of Ford trucks

Given,

P ~ Normal(580, 225)


V~ Normal 100, 64)


F~ Normal 120, 81)

also, P, V and F are independently distributed.

By the additive property of normal distributions, we have,

E(2F) = 2*120 = 240 and V(2F) = (2^2)*81 = 324

E(3V) = 3*100 = 300 and V(3V) = (9^9)*64 = 576

2F~ Normal (240, 324)
,

3VNormal (300, 576)

To compute :

P(2F > P) and P(3V > P)

Let,
$X=P-2F$
$Y=P-3V$

By the additive property of independent normal distributions (i.e. cov(F,P) and cov(v,P) are 0), we have

E(X) = 580 -240 ; V(X) = 225+349

E(Y) = 580 - 300 ; V(Y) = 225 + 576

X ~ Normal (340, 549)


Y ~ Normal(280, 801)

thus to compute,

P(X < 0) andP (y < 0)

-340 P(X < 0) Φ ) = 5.1718 * 10-4 /549
(Since
~ N(0, 1)
) obtained using R statistical software

Similary,

Ply < 0)-Φ(一200 )-2.2252 * 10-23 V801

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