Question

In a certain city, the mean price of a two liter bottle of soda is $1.50 and the standard deviation is $0.20 at various grocery stores. The mean price of one gallon of milk is $3.80 and the standard deviation is $0.30. Ten tourists came to the city togther. They separately went to randomly chosen grocery stores and each person bought one item. Six of them bought two liter bottles of soda and each of the remaining four bought one gallon of milk. i) What is the expected total amount of money that the ten people spent? ii) if the prices of individual items are independent and normally distributed, what is the probability that the total amount of money the ten people spent is at least $25?

Answer 1

1) Expected total amount of money that the ten people = $24.2

2) probability that the total amount of money the ten people spent is at least $25 = 0.3817

Explanation:

Mean Price of a two liter bottle of soda, E(x₁) = $1.50

Standard Deviation of two litre bottle of soda,
`\sigma_1 = \$ 0.20`

Mean price of one gallon of milk, E(x₂) = $3.80

Standard deviation of one gallon of milk,
`\sigma_2 = \$0.30`

Six people bought two liter bottle of soda each and four bought one gallon of milk each

Expected total amount of money that the ten people spent

`E(X) = E(6x_(1) + 4x_(2)) \\E(X) = 6E(x_(1)) + 4E(x_(2))`

E(X) = (6*1.50) + (4*3.80)

E(X) = $24.2

b) probability that the total amount of money the ten people spent is at least $25

Let us calculate the net standard deviation:

`V(X) = V(6x_(1) + 4x_(2)) \\V(X) = 6^(2) V(x_(1)) + 4^2V(x_(2))\\V(X) = 36 V(x_(1)) + 16V(x_(2))`

`V(x_(1)) = \sigma_1^(2) = 0.2^2 = 0.04\\V(x_(2)) = \sigma_2^(2) = 0.3^2 = 0.09`

V(X) = (36*0.04) + (16*0.09)

V(X) = 1.44 + 1.44 = 2.88

`\sigma = √(V(X)) \\\sigma = √(2.88) \\\sigma = 1.697`

`P( X\geq 25) = P(Z \geq (x - \mu)/(\sigma) )\\P( X\geq 25) = P(Z \geq (25 - 24.2)/(1.697) )\\P( X\geq 25) = P(Z \geq 0.4714)\\P( X\geq 25) = 0.3817`