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Chapter 7: Problem 19

According to Mediamark Research, 84 million out of 179 million adults in the United States correct their vision by using prescription eyeglasses, bifocals, or contact lenses. (Some respondents use more than one type.) What is the probability that an adult selected at random from the adult population uses corrective lenses?

Short Answer

Expert verified
The probability that a randomly selected adult from the adult population in the United States uses corrective lenses is approximately 0.46927 or 46.93%.

Step by step solution

01

Identify the given information

We are given the following information: - Total number of adults in the United States: 179 million - Number of adults using corrective lenses: 84 million
02

Calculate the probability

To find the probability that a randomly selected adult uses corrective lenses, we can use the formula: probability = \(\frac{\text{number of desired outcomes}}{\text{total number of possible outcomes}}\) In this case, the number of desired outcomes is the number of adults using corrective lenses (84 million), and the total number of possible outcomes is the total number of adults in the United States (179 million). probability = \(\frac{84 \,\text{million}}{179 \,\text{million}}\)
03

Simplify the expression

Now, we can simplify the expression and get the probability in decimal form. probability = \( \frac{84}{179}\) probability 鈮� 0.46927 The probability that a randomly selected adult from the adult population in the United States uses corrective lenses is approximately 0.46927 or 46.93%.

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Most popular questions from this chapter

In a survey of 106 senior information technology and data security professionals at major U.S. companies regarding their confidence that they had detected all significant security breaches in the past year, the following responses were obtained. $$\begin{array}{lcccc} \hline & \begin{array}{c} \text { Very } \\ \text { confident } \end{array} & \begin{array}{c} \text { Moderately } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not very } \\ \text { confident } \end{array} & \begin{array}{l} \text { Not at all } \\ \text { confident } \end{array} \\ \hline \text { Respondents } & 21 & 56 & 22 & 7 \\ \hline \end{array}$$ What is the probability that a respondent in the survey selected at random a. Had little or no confidence that he or she had detected all significant security breaches in the past year? b. Was very confident that he or she had detected all significant security breaches in the past year?

In a survey on consumer-spending methods conducted in 2006, the following results were obtained: $$\begin{array}{lccccc} \hline & & & & {\text { Debit/ATM }} & \\ \text { Payment Method } & \text { Checks } & \text { Cash } & \text { Credit cards } & \text { cards } & \text { Other } \\ \hline \text { Transactions, \% } & 37 & 14 & 25 & 15 & 9 \\ \hline \end{array}$$ If a transaction tracked in this survey is selected at random, what is the probability that the transaction was paid for a. With a credit card or with a debit/ATM card? b. With cash or some method other than with a check, a credit card, or a debit/ATM card?

Data compiled by the Department of Justice on the number of people arrested in a certain year for serious crimes (murder, forcible rape, robbery, and so on) revealed that \(89 \%\) were male and \(11 \%\) were female. Of the males, \(30 \%\) were under 18 , whereas \(27 \%\) of the females arrested were under 18 . a. What is the probability that a person arrested for a serious crime in that year was under 18 ? b. If a person arrested for a serious crime in that year is known to be under 18 , what is the probability that the person is female?

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, give an example to show why it is false. Let \(S=\left\\{s_{1}, s_{2}, \ldots, s_{n}\right\\}\) be a uniform sample space for an experiment. If \(n \geq 5\) and \(E=\left\\{s_{1}, s_{2}, s_{5}\right\\}\), then \(P(E)=3 / n\).

Let \(E\) and \(F\) be events such that \(F \subset E\). Find \(P(E \mid F)\) and interpret your result.
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