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Chapter 4: Problem 27

A random sample of 2000 adults showed that 1320 of them have shopped at least once on the Internet. What is the (approximate) probability that a randomly selected adult has shopped on the Internet?

Short Answer

Expert verified
The approximate probability that a randomly selected adult has shopped on the internet is \(\frac{1320}{2000} = 0.66 \) or 66%.

Step by step solution

01

Define the Population and Sample

Before we solve the problem, let's first identify the population and the sample. The population in this case is all adults, and the sample is the 2000 adults who were surveyed.
02

Identify the Favorable Outcome

We are interested in the adults who have shopped online. This is our event of interest, and this has occurred 1320 times out of the 2000 adults surveyed.
03

Calculate the Probability

The probability of an event is equal to the number of ways the event can occur (favorable outcomes) divided by the total number of outcomes. In our situation, the favorable outcome is the number of adults who have shopped online, and the total outcomes is the number of adults surveyed. So, the probability that a randomly selected adult has shopped online can be found by dividing 1320 by 2000.
04

Simplify the Result

Finally, simplify this result. This will give us the desired probability, which is an approximate value, due to the fact that our method of estimation assumes a simple random sample, an assumption that may not fully hold in this case.

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Most popular questions from this chapter

An ice cream shop offers 25 flavors of ice cream. How many ways are there to select 2 different flavors from these 25 flavors? How many permutations are possible?

Given that \(P(B \mid A)=.70\) and \(P(A\) and \(B)=.35\), find \(P(A)\).

A box contains 10 red marbles and 10 green marbles. a. Sampling at random from this box five times with replacement, you have drawn a red marble all five times. What is the probability of drawing a red marble the sixth time? b. Sampling at random from this box five times without replacement, you have drawn a red marble all five times. Without replacing any of the marbles, what is the probability of drawing a red marble the sixth time? c. You have tossed a fair coin five times and have obtained heads all five times. A friend argues that according to the law of averages, a tail is due to occur and, hence, the probability of obtaining a head on the sixth toss is less than \(.50 .\) Is he right? Is coin tossing mathematically equivalent to the procedure mentioned in part a or the procedure mentioned in part b above? Explain.

Given that \(A\) and \(B\) are two mutually exclusive events, find \(P(A\) or \(B\) ) for the following. a. \(P(A)=.38\) and \(P(B)=.59\) b. \(P(A)=.15\) and \(P(B)=.23\)

Find \(P(A\) or \(B)\) for the following. a. \(P(A)=.28, \quad P(B)=.39\), and \(P(A\) and \(B)=.08\) b. \(P(A)=.41, \quad P(B)=.27\), and \(P(A\) and \(B)=.19\)
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