Question

Ice cream usually comes in 1.5-quart boxes (48 fluid ounces), and ice cream scoops hold about 2 ounces. However, there is some variability in the amount of ice cream in a box as well as the amount of ice cream scooped out. We represent the amount of ice cream in the box as X and the amount scooped out as Y. Suppose these random variables have the following means, standard deviations, and variances.
mean SD variance
x 48 1 1
y 2 0.25 0.0625

(a) An entire box of ice cream, plus 3 scoops from a second box is served at a party. How much ice cream do you expect to have been served at this party? What is the standard deviation of the amount of ice cream served?
(b) How much ice cream would you expect to be left in the box after scooping out one scoop of ice cream? That is, find the expected value of X ? Y . What is the standard deviation of the amount left in the box?
(c) Using the context of this exercise, explain why we add variances when we subtract one random variable from another.

Answer 1

Answer:

(a) The expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b) The expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c) Because the variance of each variable is dependent on the other.

Step-by-step explanation:

The random variable X and Y are defined as follows:

X = amount of ice cream in the box

Y = amount of ice cream scooped out

The information provided is:

E (X) = 48

SD (X) = 1

V (X) = 1

E (Y) = 2

SD (Y) = 0.25

V (Y) = 0.0625

(a)

The total amount of ice-cream served at the party can be expressed as:

X + 3Y.

Compute the expected value of (X + 3Y) as follows:

$E(X+3Y)=E(X)+3E(Y)\\= 48+(3\times2)\\=48+6\\=54$

Compute the variance of (X + 3Y) as follows:

$V(X+3Y) = V (X)+3^{2}V(Y)+2\times 3Cov (X,Y)\\=1+(9\times0.0625)+0\\=1.5625$

Then the standard deviation of (X + 3Y) is:

$SD(X + 3Y) =\sqrt{V(X + 3Y)}\\\sqrt{1.5625}\\=1.25$

Thus, the expected value and standard deviation of the amount of ice cream served at the party are 54 ounces and 1.25 ounces respectively.

(b)

The amount of ice-cream left in the box after scooping out one scoop is represented as follows:

X - Y.

Compute the expected value of (X - Y) as follows:

$E(X-Y)=E(X)-E(Y)\\=48-2\\=46$

Compute the variance of (X - Y) as follows:

$V(X - Y) =V(X)+V(Y)-2Cov(X,Y)\\=1+0.0625-0\\=1.0625$

Then the standard deviation of (X - Y) is:

$SD(X-Y) =\sqrt{V(X -Y)}\\\sqrt{1.0625}\\=1.031$

Thus, the expected value and standard deviation of the amount of ice cream left in the box after scooping out one scoop are 46 ounces and 1.031 ounces respectively.

(c)

The variance of the sum or difference of two variables is computed by adding the individual variances. This is because the variance of each variable is dependent on the others.