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Worked solution

Calculate expected car-wash earnings

Let XX be the number of cars passing through a car wash between 4:00 pm and 5:00 pm on a sunny Friday, with probability distribution

xx456789
P(X=x)P(X=x)1/121/121/121/121/41/41/41/41/61/61/61/6

The manager pays the attendant Y=2X−1Y=2X-1 dollars. Find the attendant’s expected earnings for this time period.

CJC · Lecture · Example 10

Method 1: average the function directly

Use E(Y)=E[g(X)]=∑xg(x)f(x)E(Y)=E[g(X)]=\sum_xg(x)f(x). Here E(Y)=E(2X−1)=∑x=49(2x−1)P(X=x)E(Y)=E(2X-1)=\sum_{x=4}^{9}(2x-1)P(X=x). Expanding every source term gives [2(4)−1](112)+[2(5)−1](112)+[2(6)−1](14)+[2(7)−1](14)+[2(8)−1](16)+[2(9)−1](16)=38/3[2(4)-1](\frac1{12})+[2(5)-1](\frac1{12})+[2(6)-1](\frac14)+[2(7)-1](\frac14)+[2(8)-1](\frac16)+[2(9)-1](\frac16)=38/3.

How did this part go?

Method 2: form the transformed distribution

The map y=2x−1y=2x-1 is one-to-one on the six listed values, so each transformed value keeps its original mass. Retain every row of the source table:

xx456789
P(X=x)P(X=x)1/121/121/121/121/41/41/41/41/61/61/61/6
y=2x−1y=2x-17911131517
P(Y=y)P(Y=y)1/121/121/121/121/41/41/41/41/61/61/61/6
yP(Y=y)yP(Y=y)7/127/129/129/1211/411/413/413/415/615/617/617/6

E(Y)=7/12+9/12+11/4+13/4+15/6+17/6=38/3E(Y)=7/12+9/12+11/4+13/4+15/6+17/6=38/3 dollars, or approximately 12.67 dollars (supplementary rounding).

How did this part go?

Common mistakes

Average the transformed amounts using their probabilities. Keeping individual masses works here because the transformation is one-to-one; a many-to-one transformation requires combining masses as in Example 5.

Read the hint