Practise calculating or recovering means and standard deviations for normal sums by using linear expectation and independent variance relationships.
Each full carton of Grade A eggs consists of 1 randomly selected empty cardboard container and 12 randomly selected eggs. The weights of such full cartons are approximately normally distributed with a mean of 840 grams and a standard deviation of 7.9 grams.
Official holistic scoring rubric
Parts (a), (b-i), and (b-ii) are each classified as essentially correct (E), partially correct (P), or incorrect (I). Minor arithmetic errors are not penalized.
Part (a): E requires all three components: use of a normal distribution with the correct parameter values, the correct boundary value, and the correct normal probability consistent with those choices. P requires exactly two components. Otherwise the part is I. A statistical-notation error lowers the part by one level, and acceptable correct probabilities range from 0.1020 through 0.1038.
Part (b-i): E requires a correct expected-value equation and correct solution, or another correct numerical procedure for the expected weight of one egg. P applies when the correct procedure is indicated but expected-value properties are applied incorrectly, or when the procedure is communicated poorly. Otherwise the part is I.
Part (b-ii): E requires combining variances and all three components: subtracting variances, correctly using 12 in the calculation, and reporting the corresponding correct standard deviation. P requires combining variances and exactly two components. Otherwise the part is I.
Overall score: 4 = all three parts E. 3 = two parts E and one part P, or one part E and two parts P. 2 = two parts E and one part I, one part E with one part P and one part I, or all three parts P. 1 = one part E and two parts I, or one or two parts P.
The weights of the empty cardboard containers have a mean of 20 grams and a standard deviation of 1.7 grams. It is reasonable to assume independence between the weights of the empty cardboard containers and the weights of the eggs. It is also reasonable to assume independence among the weights of the 12 eggs that are randomly selected for a full carton.
Let the random variable X be the weight of a single randomly selected Grade A egg.
What is the mean of X ?
Let W represent the weight of a randomly selected full carton of eggs, P the weight of the packaging, and Xi the weight of the i th egg, for i=1,2…,12.
Note that W=P+X1+X2+…+X12.
Properties of expected values establish that E(W)=E(P)+E(X1)+…+E(X12).
Because all 12 eggs have the same mean weight, this becomes E(W)=E(P)+12×E(Xi).
We were told that E(W)=840 and E(P)=20, so we can solve
What is the standard deviation of X ?
Because of independence, properties of variance establish that
Var(W)=Var(P)+Var(X1)+Var(X2)+…+Var(X12).
Because all 12 eggs have the same variance of their weights, this becomes
Var(W)=Var(P)+12×Var(Xi).
We were told that SD(W)=7.9 and SD(P)=1.7. Therefore, Var(W)=(7.9)2=62.41 and Var(P)=(1.7)2=2.89.
We can solve 62.41=2.89+12×Var(Xi) to find Var(Xi)=1262.41−2.89=4.96. Thus, SD(Xi)=(4.96)≈2.23 grams.