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The table gives the probability distribution of the number of books sold in a day at a bookstore. What is the probability of 16 or more books being sold on a given day?

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|}
\hline
number of books sold in a day & 0-5 & 6-10 & 11-15 & 16-20 & 21-25 \\
\hline
probability & 0.110 & 0.206 & 0.464 & 0.201 & ? \\
\hline
\end{tabular}
\][/tex]

A. 0.220
B. 0.201
C. 0.464
D. 0.019



Answer :

GinnyAnswer
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.

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