To find out how many books of each subject the librarian likely has based on the frequency of each subject and the total number of books (28), we first need to determine the total frequency and then the probability of selecting each subject.
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Calculate the total frequency: \[ \text{Total Frequency} = 40 + 16 + 20 + 36 = 112 \]
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Calculate the probability for each subject:
- Science: \[ P(\text{Science}) = \frac{40}{112} \]
- Math: \[ P(\text{Math}) = \frac{16}{112} \]
- History: \[ P(\text{History}) = \frac{20}{112} \]
- Reading: \[ P(\text{Reading}) = \frac{36}{112} \]
- Normalize the probabilities for simplicity (as fractions):
- Science: \[ P(\text{Science}) = \frac{40}{112} = \frac{10}{28} \]
- Math: \[ P(\text{Math}) = \frac{16}{112} = \frac{4}{28} \]
- History: \[ P(\text{History}) = \frac{20}{112} = \frac{5}{28} \]
- Reading: \[ P(\text{Reading}) = \frac{36}{112} = \frac{9}{28} \]
- Use these probabilities to calculate the expected number of books for each subject in a total of 28 books:
- Science: \[ N_{\text{Science}} = 28 \times \frac{10}{28} = 10 \]
- Math: \[ N_{\text{Math}} = 28 \times \frac{4}{28} = 4 \]
- History: \[ N_{\text{History}} = 28 \times \frac{5}{28} = 5 \]
- Reading: \[ N_{\text{Reading}} = 28 \times \frac{9}{28} = 9 \]
From the calculations, we find that the likely number of books for each subject is:
- Science: 10
- Math: 4
- History: 5
- Reading: 9
This matches with option A: 10 science books, 4 math books, 5 history books, and 9 reading books.