Problem 24 - Basic Statistics for Business and Economics with Student

Question

Backyard Retreats, Inc., sells an exclusive line of pools, hot tubs, and spas. It is located just off the Bee Line Expressway in Orlando, Florida. The owner reports 20 percent of the customers entering the store will make a purchase of at least \(\$ 50 .\) Suppose 15 customers enter the store before 10 a.m. on a particular Saturday. a. How many of these customers would you expect to make a purchase of at least \(\$ 50 ?\) b. What is the probability exactly five of these customers make a purchase of at least \(\$ 50 ?\) c. What is the probability at least five of these customers make a purchase of at least \(\$ 50 ?\) d. What is the probability at least one customer makes a purchase of at least \(\$ 50 ?\)

Step-by-Step Solution

Answer

a. 3 customers; b. 0.103; c. 0.1074; d. 0.9648.

Step 1: Understanding the Problem

We know that each customer has a 20% chance of making a purchase of at least $50. The situation can be modeled using a binomial distribution because we have a fixed number of trials (customers), two possible outcomes for each trial (purchase or not), and the probability of success (purchase) is constant at 0.20. The number of trials is 15 since 15 customers enter the store.

Step 2: Calculating Expected Number of Purchases

To calculate the expected number of customers making a purchase, use the formula for the expected value in a binomial distribution: \[ E(X) = n imes p \]where \( n = 15 \) customers, and \( p = 0.20 \) probability of purchase.\[ E(X) = 15 imes 0.20 = 3 \] We expect 3 customers to make a purchase of at least $50.

Step 3: Probability of Exactly Five Purchases

To find the probability of exactly 5 customers making a purchase, we use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]where \( n = 15 \), \( p = 0.20 \), and \( k = 5 \).\[ P(X = 5) = \binom{15}{5} (0.20)^5 (0.80)^{10} \]Calculate \( \binom{15}{5} = 3003 \),then \( (0.20)^5 = 0.00032 \),and \( (0.80)^{10} = 0.1074 \).Thus, \[ P(X = 5) = 3003 \times 0.00032 \times 0.1074 \approx 0.103 \].

Step 4: Probability of At Least Five Purchases

To find the probability of at least 5 customers making a purchase, we sum the probabilities from 5 to 15:\[ P(X \geq 5) = 1 - P(X < 5) \]Calculating probabilities for 0 to 4 and summing them, we get \( P(X < 5) \approx 0.8926 \) (using cumulative binomial probability tables or calculator). Therefore,\[ P(X \geq 5) = 1 - 0.8926 = 0.1074 \].

Step 5: Probability of At Least One Purchase

To find the probability that at least one customer makes a purchase, use:\[ P(X \geq 1) = 1 - P(X = 0) \].Calculate \( P(X = 0) \) using the binomial formula:\[ P(X = 0) = \binom{15}{0} (0.20)^0 (0.80)^{15} = 1 \times 1 \times 0.80^{15} \approx 0.0352 \].Thus,\[ P(X \geq 1) = 1 - 0.0352 = 0.9648 \].

Key Concepts

Probability

Probability is a crucial concept in statistics that measures how likely an event is to occur. In our problem, each customer has a 20% chance of making a purchase worth at least $50. This scenario is a classic example of a binomial distribution, characterized by items like:

• A fixed number of trials: 15 customers visiting the store.
• Two possible outcomes: a customer makes a purchase or does not.
• A constant probability of success: 20% or 0.20 likelihood of making a purchase.
• Independent events: Each customer's decision to purchase does not impact others.

To understand further, we can visualize the probability as calculating the likelihood for each number of purchases, from 0 up to 15, using binomial probability formulas.

Expected Value

The expected value in a probability scenario gives us a sense of the average or mean number of successes we could anticipate in the long run. It's like asking: if today was any normal day, how many people would actually buy something worth at least $50? In our problem, using the formula for expected value in binomial distribution, \[ E(X) = n \times p \] where \( n = 15 \) and \( p = 0.20 \), the expected number of customers making a purchase is 3. Expected value doesn't mean exactly 3 will purchase every time, but rather over many similar Saturdays, we'd average about 3 purchasing customers. This paints a bigger picture of what we expect on such shopping days.

Cumulative Distribution

Cumulative distribution helps us understand the probability of a range of outcomes rather than a single result. For the store scenario, we used it to calculate "at least" probabilities, like the probability that at least 5 customers make a purchase. The key here is to calculate the cumulative distribution, which is the probability of 0 through 4 customers purchasing and subtract from 1 to find probabilities of 5 or more having bought something: \[ P(X \geq 5) = 1 - P(X < 5) \] This method helps us avoid summing multiple individual probabilities and provides a comprehensive view of likelihood across a range of outcomes.

Statistical Analysis

Statistical analysis involves using principles and calculations to interpret real-life data and make informed predictions and decisions. Through statistical analysis, such as calculating probabilities and expected values, businesses like Backyard Retreats, Inc. can better forecast their sales and prepare accordingly. Statistical methods give insights into customer behavior by providing facts and figures about expected sales volume, the probability of multiple customers making purchases, and the likelihood of at least one sale. This deeper understanding allows the business to optimize inventory, improve customer service, and strategize marketing efforts, helping to maximize potential profits and efficiency.