Problem 43
Question
Recent statistics suggest that \(15 \%\) of those who visit a retail site on the Internet make a purchase. A retailer wished to verify this claim. To do so, she selected a sample of 16 "hits" to her site and found that 4 had actually made a purchase. a. What is the likelihood of exactly four purchases? b. How many purchases should she expect? c. What is the likelihood that four or more "hits" result in a purchase?
Step-by-Step Solution
Verified Answer
a. Likelihood of exactly four purchases is ~0.1474. b. Expect about 2.4 purchases. c. Likelihood for four or more is ~0.3064.
1Step 1: Define the Parameters
This is a binomial distribution problem where we have \( n = 16 \) (the number of trials, or site "hits"), and the probability of success (a purchase) is \( p = 0.15 \).
2Step 2: Calculate Likelihood of Exactly Four Purchases
Use the binomial probability formula: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) where \( k = 4 \). Thus, calculate \( P(X = 4) = \binom{16}{4} (0.15)^4 (0.85)^{12} \). The calculation gives \( P(X = 4) \approx 0.1474 \).
3Step 3: Calculate Expected Number of Purchases
The expected number of successes (purchases) in a binomial distribution is given by \( E(X) = n \times p \). Here it is \( E(X) = 16 \times 0.15 = 2.4 \).
4Step 4: Calculate Likelihood of Four or More Purchases
Calculate \( P(X \geq 4) = 1 - P(X < 4) \). First find \( P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \) using the binomial formula for each case, then subtract this sum from 1. The resulting probability is \( P(X \geq 4) \approx 0.3064 \).
Key Concepts
Probability of SuccessExpected Number of PurchasesProbability of at Least EventBinomial Probability Formula
Probability of Success
The probability of success in a binomial distribution is a key factor to understand when calculating probabilities in such scenarios. In the context of our exercise, "success" means a website visitor makes a purchase. The given probability of a customer making a purchase is stated to be 15%, or mathematically, \( p = 0.15 \). This probability is determined based on historical data or expected statistics. It helps us evaluate the likelihood of different numbers of "successes" (purchases) during multiple trials or "attempts." With each attempt (each hit on the website), there is a constant probability \( p \) of achieving a success, which is vital for the binomial probability calculations to follow. Understanding this probability aids us in predicting and analyzing the purchasing behavior of visitors and ultimately helps improve business strategies.
Expected Number of Purchases
The expected number of purchases is an average estimate of how many successes, or purchases, we might expect in our set of trials. In a binomial distribution, the formula to find this expectation is \( E(X) = n \times p \). Here, \( n \) represents the total number of trials (in this case, site hits), which is 16; and \( p \) represents the probability of success on each trial, or 0.15. Substituting these values, we calculate:
- \( E(X) = 16 \times 0.15 = 2.4 \)
Probability of at Least Event
Calculating the probability of at least a certain number of successes is a common scenario. In our exercise, we seek the probability that four or more website hits result in a purchase. This is expressed as \( P(X \geq 4) \). Instead of computing this directly for all cases of \( k = 4 \) and greater, we use a complementary approach. First, compute \( P(X < 4) \)—the probability of fewer than 4 purchases—and then subtract this from 1, since probabilities sum to 1.
- \( P(X \geq 4) = 1 - P(X < 4) \)
Binomial Probability Formula
The binomial probability formula is essential to find the likelihood of obtaining exactly \( k \) successes in \( n \) trials. The formula is:\[P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\]This formula includes:
- \( \binom{n}{k} \) which is the binomial coefficient, calculating the number of ways \( k \) successes can occur in \( n \) trials.
- \( p^k \) which accounts for the probability of getting exactly \( k \) successes.
- \( (1-p)^{n-k} \) which represents the probability of the remaining \( n-k \) trials resulting in failure.
- \( P(X = 4) = \binom{16}{4} \times (0.15)^4 \times (0.85)^{12} \)
Other exercises in this chapter
Problem 41
A federal study reported that \(7.5 \%\) of the U.S. workforce has a drug problem. A drug enforcement official for the state of Indiana wished to investigate th
View solution Problem 42
The Bank of Hawaii reports that \(7 \%\) of its credit card holders will default at some time in their life. The Hilo branch just mailed out 12 new cards today.
View solution Problem 44
Acceptance sampling is a statistical method used to monitor the quality of purchased parts and components. To ensure the quality of incoming parts, a purchaser
View solution Problem 45
Unilever Inc. recently developed a new body wash with a scent of ginger. Their research indicates that \(30 \%\) of men like the new scent. To further investiga
View solution