To determine the probability of selling 16 or more books in a given day, we need to consider the probabilities for the categories "16-20 books" and "21-25 books."
The given probabilities are:
- 0-5 books: 0.110
- 6-10 books: 0.206
- 11-15 books: 0.464
- 16-20 books: 0.201
- 21-25 books: ?
First, we recognize that the total probability for all possible outcomes should sum to 1. Therefore, we can calculate the missing probability for the "21-25 books" category.
We can use the equation:
[tex]\[ \text{Total Probability} = \text{P(0-5)} + \text{P(6-10)} + \text{P(11-15)} + \text{P(16-20)} + \text{P(21-25)} \][/tex]
Substituting the given values:
[tex]\[ 1 = 0.110 + 0.206 + 0.464 + 0.201 + \text{P(21-25)} \][/tex]
Solving for P(21-25):
[tex]\[ \text{P(21-25)} = 1 - (0.110 + 0.206 + 0.464 + 0.201) \][/tex]
[tex]\[ \text{P(21-25)} = 1 - 0.981 \][/tex]
[tex]\[ \text{P(21-25)} = 0.019 \][/tex]
Now, we need to find the probability of selling 16 or more books. This is the sum of the probabilities for the "16-20 books" and "21-25 books" categories:
[tex]\[ \text{P(16 or more)} = \text{P(16-20)} + \text{P(21-25)} \][/tex]
[tex]\[ \text{P(16 or more)} = 0.201 + 0.019 \][/tex]
[tex]\[ \text{P(16 or more)} = 0.220 \][/tex]
Thus, the probability of selling 16 or more books in a given day is:
[tex]\[ \boxed{0.220} \][/tex]
The correct answer is A. 0.220.
