Question

# Suppose that you are a student-employee in the Accounting Department of the Business School and they...

Suppose that you are a student-employee in the Accounting Department of the Business
School and they agree to pay you using the Random Pay system. Each week the Chair
flips a coin. If it comes up heads, your pay for the week is \$80; if it comes up tails your
pay for the week is \$40. Your friend is working for the Engineering Department and
makes \$65 per week. If your total earning in 100 weeks is a random variable that
follows a Normal probability distribution, the probability that your total earnings in 100
weeks are more than hers is approximately:

Total earning of friend in 100 weeks is given as: 65*100 = 6500

For us, the probability distribution of earnings in a week is given as:
P(X = 80) = P(X = 40) = 0.5

E(X) = 0.5*(80 + 40) = 60

E(X2) = 0.5*(802 + 402) = 4000
Therefore, Var(X) = E(X2) - [E(X)]2 = 4000 - 602 = 400

Therefore the distribution for 100 weeks mean earning is given here as: (Using Central limit theorem)

The required probability now is computed here as:

Converting it to a standard normal variable, we get here:

Getting it from the standard normal tables, we get here:

Therefore 0.0062 is the required probability here.

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