Question

the national school lunch program took a random sample of students who ordered school lunch. use the partially completed relative frequency table given to find the percentage of students who had cheese pizza, given that they had chocolate milk.
(table with rows: fruit juice, chocolate milk, total; columns: cheese pizza, chicken fingers, total. entries: cheese pizza - fruit juice: 33%, total: 55%; chicken fingers - total: 45%; total - fruit juice: 60%, total: 100%. multiple choice options: 22%, 44%, 55%, 82%)

Explanation:

Step1: Recall Conditional Probability Formula

We need to find \( P(\text{Cheese Pizza} | \text{Chocolate Milk}) \). By the formula for conditional probability, \( P(A|B)=\frac{P(A\cap B)}{P(B)} \). In terms of percentages from the table, \( P(\text{Cheese Pizza and Chocolate Milk}) = 55\% \) (wait, no, let's check the table again. Wait, the table has:

Rows: Fruit Juice, Chocolate Milk, Total

Columns: Cheese Pizza, Chicken Fingers, Total

From the table:

- For Cheese Pizza: Total (row) is 55% (wait, no, the "Total" column for Cheese Pizza is 55%? Wait, the table:

Cheese Pizza row (wait, no, columns are Cheese Pizza, Chicken Fingers, Total; rows are Fruit Juice, Chocolate Milk, Total.

Wait, the "Total" row (last row) has Cheese Pizza: 55%, Chicken Fingers: 45%, Total: 100%.

The "Fruit Juice" row: Cheese Pizza: 33%, Total (Fruit Juice row): 60%.

We need \( P(\text{Cheese Pizza} | \text{Chocolate Milk}) \). So \( A = \text{Cheese Pizza} \), \( B = \text{Chocolate Milk} \).

First, find \( P(A\cap B) \) (percentage of students who had both Cheese Pizza and Chocolate Milk) and \( P(B) \) (percentage of students who had Chocolate Milk).

From the table, the "Total" column for Cheese Pizza is 55% (total who had Cheese Pizza), Chicken Fingers is 45% (total who had Chicken Fingers). The "Fruit Juice" row total is 60%, so the "Chocolate Milk" row total is \( 100\% - 60\% = 40\% \)? Wait, no, maybe the table is structured as:

Columns: Cheese Pizza (CP), Chicken Fingers (CF), Total (Column Total)

Rows: Fruit Juice (FJ), Chocolate Milk (CM), Total (Row Total)

So:

- FJ row: CP = 33%, CF =?, FJ row total = 60%

- CM row: CP =?, CF =?, CM row total =?

- Total row: CP = 55%, CF = 45%, Total = 100%

- FJ row total = 60%, so CM row total = 100% - 60% = 40%

Now, column total for CP is 55% (so CP in FJ + CP in CM = 55%). FJ row CP is 33%, so CM row CP = 55% - 33% = 22%? Wait, no, maybe the table is a relative frequency table where each cell is the percentage of the total sample. Wait, the problem is to find the percentage of students who had cheese pizza, given that they had chocolate milk. So it's a conditional probability: \( P(CP | CM) = \frac{\text{Number of students with CP and CM}}{\text{Number of students with CM}} \times 100\% \)

From the table, the "Total" row (row total) for CP is 55% (so 55% of all students had CP), CF is 45% (45% had CF). The "FJ" row total is 60% (60% had FJ), so the "CM" row total is \( 100\% - 60\% = 40\% \) (40% had CM).

Now, the "CP" column total is 55%, which is FJ row CP (33%) + CM row CP (let's call it x). So \( 33\% + x = 55\% \) ⇒ \( x = 55\% - 33\% = 22\% \). So the number of students with both CP and CM is 22% (of total sample), and the number of students with CM is 40% (of total sample). Wait, no, that can't be, because the "CM" row total is 40%, so if CM row CP is 22%, then CM row CF is \( 40\% - 22\% = 18\% \).

But wait, the "Total" column for CF is 45%, which is FJ row CF + CM row CF. FJ row CF is \( 60\% - 33\% = 27\% \), so CM row CF is \( 45\% - 27\% = 18\% \), which matches \( 40\% - 22\% = 18\% \). Good.

Now, \( P(CP | CM) = \frac{\text{CM row CP}}{\text{CM row total}} = \frac{22\%}{40\%} \)? Wait, no, that's not right. Wait, maybe the table is a joint relative frequency table where each cell is the percentage of the total sample. Wait, the "Total" row (row total) for CP is 55% (so 55% of all students had CP), CF is 45% (45% had CF). The "FJ" row total is 60% (60% had FJ), so "CM" row total is 40% (40% had CM). The "CP" column: FJ row is 33% (33% of total had FJ and CP), so CM row CP is \( 55\% - 33\% = 22\% \) (22% of total had CM and CP). Then, the percentage of students who had CM is 40% (since FJ is 60%, total is 100%). So \( P(CP | CM) = \frac{22\%}{40\%} = 55\% \)? Wait, no, 22 divided by 40 is 0.55, which is 55%? Wait, but the options are 22%, 44%, 55%, 82%. Wait, maybe I misread the table.

Wait, the table given in the image:

Columns: Cheese Pizza, Chicken Fingers, Total

Rows: Fruit Juice, Chocolate Milk, Total

Cheese Pizza column total: 55%

Chicken Fingers column total: 45%

Fruit Juice row: Cheese Pizza = 33%, Total (Fruit Juice row) = 60%

Total row: 100%

So Chocolate Milk row total is \( 100\% - 60\% = 40\% \)

Cheese Pizza in Chocolate Milk row: \( 55\% - 33\% = 22\% \) (since total Cheese Pizza is 55%, 33% from Fruit Juice, so 22% from Chocolate Milk)

Now, \( P(CP | CM) = \frac{\text{Cheese Pizza and Chocolate Milk}}{\text{Chocolate Milk total}} = \frac{22\%}{40\%} \)? No, that's 55%? Wait, 22 divided by 40 is 0.55, which is 55%. Wait, but 22% is the percentage of total sample who had both, and 40% is the percentage of total sample who had CM. So 22/40 = 0.55 = 55%. Wait, but the options include 55%. Wait, maybe the table is structured differently. Alternatively, maybe the "Total" column for Cheese Pizza is 55% (of the Chocolate Milk row)? No, the problem says "given that they had chocolate milk", so we look at the Chocolate Milk row.

Wait, maybe the table is:

- Cheese Pizza: Fruit Juice = 33%, Chocolate Milk =?, Total = 55%

- Chicken Fingers: Fruit Juice =?, Chocolate Milk =?, Total = 45%

- Total (Fruit Juice row) = 60%, Total (Chocolate Milk row) = 40%, Total (Total row) = 100%

So Cheese Pizza in Chocolate Milk row: 55% - 33% = 22%

Chicken Fingers in Fruit Juice row: 60% - 33% = 27%

Chicken Fingers in Chocolate Milk row: 45% - 27% = 18%

Now, the Chocolate Milk row: Cheese Pizza = 22%, Chicken Fingers = 18%, Total = 40% (22+18=40, correct)

Now, \( P(CP | CM) = \frac{\text{CP in CM row}}{\text{CM row total}} = \frac{22\%}{40\%} = 55\% \)? Wait, 22 divided by 40 is 0.55, which is 55%. So the answer is 55%.

Step2: Verify with Conditional Probability

Conditional probability formula: \( P(A|B) = \frac{P(A\cap B)}{P(B)} \)

Here, \( A = \text{Cheese Pizza} \), \( B = \text{Chocolate Milk} \)

\( P(A\cap B) \) is the percentage of students who had both, which is the Cheese Pizza value in the Chocolate Milk row: 22% (of total sample)

\( P(B) \) is the percentage of students who had Chocolate Milk, which is the total of the Chocolate Milk row: 40% (of total sample)

Wait, no, that can't be, because 22% is the number of students who had both, and 40% is the number who had CM. So 22/40 = 0.55 = 55%. So the percentage is 55%.

Answer:

55% (corresponding to the 55% option)