Question

Betting on dinner, Part II. Exercise 4.24 introduces a promotion at a restaurant where prices of menu items are determined randomly following some underlying distribution. We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all ?ve baskets of fries?

Answer 1

X: price of basket of fries

X follows a normal distribution with mean 6 standard deviation of 2

Y : Price of 5 baskets of fries

Y = 5X

As X follows a normal distribution with mean 6 standard deviation of 2, then Y: 5X also follows normal distribution with mean : 5 x 6 =30 and standard deviation 5 x 2=10

Amount available in your pocket : $28

You have Enough money if (5X) the price of the 5 baskets of fries is less than 28 i.e 5X<28

probability that you would have enough money to pay for all five baskets of fries = P(5X < 28) =P(Y<28)

Z-score for 28 = (28-30)/10 = -2/10 = -0.2

From Standard normal tables :

P(Z < -2) = 0.4207

probability that you would have enough money to pay for all five baskets of fries = P(5X < 28) =P(Y<28) = 0.4207

probability that you would have enough money to pay for all five baskets of fries = 0.4207