Question

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:

Answer #1

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(X^{2}) = 0.5*(80^{2} + 40^{2}) =
4000

Therefore, Var(X) = E(X^{2}) - [E(X)]^{2} = 4000 -
60^{2} = 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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