Question

Suppose that the distribution of the weights of bags of carrots from brand A is N(1.2,0.049) and the distribution of the weights of bags of carrots from brand A is N(3.5, 0.081). The weights of bags from two brands is independent. Selecting bags at random find

a) The probability that the sum of a random sample of the weights of three bags from brand A exceeds the weight of a bag from brand B. Give answer to the 4th decimal.

b) The probability that the weight of a bag of carrots from brand A or one-third the weight of a bag of carrots from brand B is at least 1.2 pounds. Give answer to the 4th decimal.

Answer 1

Here the distribution of weights of bags of carrots from brand A is N (1.2 , 0.049)

if x is the weight of a random bag of carrots than x ~ N(1.2, 0.049)

if y is the weight of a random bag of carrots from brand B is N (3.5 , 0.081)

y ~ N (3.5 , 0.081)

(a) Here distribution of sum of a random sample of the weights of three bags from brand A is N (1.2 * 3 , 0.049 * 3) or

X ~ N(3.6 , 0.147)

where X is the sum of of a random sample of the weights of three bags from brand A

so here if z = X - y

so the distribution of Z would be normal

where E[Z] = E[X - y] = E[X] - E[y] = 3.6 - 3.5 = 0.1

Var[Z] = Var[X- y] = Var [X] + Var[Y] = 0.147 + 0.081 = 0.228

SD[Z] = sqrt(0.228) = 0.4775

so here we have to find

Pr(z > 0) =1 - Pr(z < 0) =1 - Pr(z < 0)

Pr(z < 0) = NORMDIST(z < 0 ; 0.1 ; 0.4775)

Z value = (0 - 0.1)/0.4775 = -0.2094

Pr(z > 0) =1 - Pr(z < 0) =1 - Pr(z < 0) = 1-   (-0.2094)

where is the standard normal cumulative distribution.

Pr(z > 0) =1 - Pr(z < 0) =1 - Pr(z < 0) = 1-   (-0.2094) = 1- 0.4171 = 0.5829

(B) The probability that the weight of a bag of carrots from brand A  is at least 1.2 pounds.

Pr(x > 1.2 pounds) = 1 - Pr(x < 1.2 pounds; 1.2 pounds ; 0.049) = 0.5

probability that one-third the weight of a bag of carrots from brand B is at least 1.2 pounds

where if Y is the one third the weight of a bag of carrots from brand B.

E[Y] = 3.5/3 = 1.16667

VaR[Y] = 0.081/3 = 0.027

SD[Y] = 0.1643

so, here

Pr(Y > 1.2 pounds) = 1 - Pr(Y < 1.2 pounds; 1.1667 pounds; 0.1643 pounds)

Z = (1.2 - 1.1667)/0.1643 = 0.2027

Pr(Y > 1.2 pounds) = 1 - Pr(Y < 1.2 pounds; 1.1667 pounds; 0.1643 pounds) = 1 - Pr(Z < 0.2027) = 1 - 0.5804 = 0.4196

so

Now these two events are independent.

so Pr(the weight of a bag of carrots from brand A or one-third the weight of a bag of carrots from brand B > 1.2 pounds)

= Pr(x > 1.2 pounds) + Pr(Y > 1.2 pounds) - Pr(x > 1.2 pounds) * Pr(Y > 1.2 pounds)

= 0.5 + 0.4196 - 0.5 * 0.4196 = 0.7098