Question

An article in American Demographics claims that more than twice as many shoppers are out shopping on the weekends than during the week. \(^{3}\) Not only that, such shoppers also spend more money on their purchases on Saturdays and Sundays! Suppose that the amount of money spent at shopping centers between 4 p.m. and 6 P.m. on Sundays has a normal distribution with mean \(\$ 85\) and with a standard deviation of \(\$ 20\). A shopper is randomly selected on a Sunday between 4 p.m. and 6 p.m. and asked about his spending patterns. a. What is the probability that he has spent more than \(\$ 95\) at the mall? b. What is the probability that he has spent between \(\$ 95\) and \(\$ 115\) at the mall? c. If two shoppers are randomly selected, what is the probability that both shoppers have spent more than \(\$ 115\) at the mall?

Short answer

Answer: The probability that both shoppers have spent more than $115 at the mall is approximately 0.0045.

Step-by-step solution

Calculate the z-score for each given spending amount

To find the z-score, use the formula: \(z = \frac{X - \mu}{\sigma}\) Where \(X\) is the specific spending amount, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. For part (a): \(z_1 = \frac{95 - 85}{20} = \frac{10}{20} = 0.5\) For part (b): \(z_2 = \frac{115 - 85}{20} = \frac{30}{20} = 1.5\) Now we have the z-scores for each part of the question.

Find the respective probabilities

Using a z-table or an online z-score calculator, determine the probabilities for each z-score obtained in Step 1. For part (a): The probability of a shopper spending more than \(\$95\) at the mall is \(P(z > 0.5) = 1 - P(z\le 0.5)\). Find the value for \(P(z\le 0.5)\) using a z-table or calculator, which yields \(0.6915\). Therefore, the probability of a shopper spending more than \(\$95\) is \(1 - 0.6915 = 0.3085\). For part (b): The probability of a shopper spending between \(\$95\) and \(\$115\) at the mall is \(P(0.5 < z < 1.5) = P(z\le 1.5) - P(z\le 0.5)\). Using a z-table or calculator, we find the probability values for \(P(z\le 1.5) = 0.9332\) and \(P(z\le 0.5) = 0.6915\). Then, calculate the difference to find the probability: \(0.9332 - 0.6915 = 0.2417\).

Calculate the probability for selecting two shoppers

For part (c): As the two shoppers are randomly selected, we must assume the spending habits are independent events. Therefore, to find the probability both shoppers have spent more than \(\$115\) at the mall, calculate the probability for one shopper and then square it. First, find the z-score for \(\$115\): \(z = \frac{115 - 85}{20} = 1.5\). Then, calculate the corresponding probability: \(P(z > 1.5) = 1 - P(z\le 1.5)\). Using a z-table or calculator, we find the probability value for \(P(z\le 1.5) = 0.9332\). Then, calculate the probability of a shopper spending more than \(\$115\): \(1 - 0.9332 = 0.0668\). Finally, square the probability calculated for one shopper: \((0.0668)^2 = 0.0045\).

Problem Summary

Here are the probabilities for each part of the question: a. The probability that a shopper has spent more than \(\$95\) at the mall is approximately \(0.3085\). b. The probability that a shopper has spent between \(\$95\) and \(\$115\) at the mall is approximately \(0.2417\). c. The probability that both shoppers have spent more than \(\$115\) at the mall is approximately \(0.0045\).

Key concepts

Probability Calculation

Probability calculation is a key component of statistics that helps in determining the likelihood of an event occurring. In the context of our shopping exercise, we want to find out the probability of a shopper spending a specific amount at the mall. This involves calculating the area under the normal distribution curve for different scenarios.

Normal distribution is often used in such calculations because of its bell-shaped curve, which represents many naturally occurring variables. The mean of the distribution tells us the average spending, while the standard deviation indicates the spread of data around this mean. By converting spending amounts into standard scores, or z-scores, we can easily use the standard normal distribution to find probabilities.

To determine the desired probability, you will often refer to z-tables, which list probabilities associated with particular z-scores. These tables, or their digital equivalents, are extensive but allow for quick lookup to see the probability of an event occurring under a standard normal distribution.

Z-score

A z-score is an essential concept in statistics that allows you to compare different data points relative to the mean of a distribution. It gives insight on how many standard deviations a particular data point is from the mean. This measurement helps in assessing probabilities within a normal distribution.

The formula to compute a z-score is simple: \[ z = \frac{X - \mu}{\sigma} \]where:

• \(X\) is the specific data value or spending amount in our exercise,
• \(\mu\) is the mean of the distribution, which in this case is the average amount shoppers spend,
• \(\sigma\) is the standard deviation.

Z-scores are useful because they place different data points on a standard scale, allowing for an easy comparison across different scenarios. They enable you to consult z-tables for the associated probability, connecting individual data points back to a bigger picture within a normal distribution. This way, we can find out the likelihood of spending more than a certain amount or between different amounts by referring to these standardized scores.

Spending Patterns

Understanding spending patterns, particularly in the context of normal distribution, requires examining how much consumers are likely to spend within a given period. This involves looking at the average spending amount (mean) and how much individual amounts typically deviate (standard deviation) from this average.

Spending patterns on weekends can differ significantly from weekday behavior. This exercise focuses on a specific period, between 4 p.m. and 6 p.m. on Sundays, where spending more at malls appears more common. Analyzing such patterns helps businesses plan promotions or understand peak sales times.

• **Mean (Average Spending)**: In our example, the mean is $85. This tells us that most shoppers, on average, spend around this amount.
• **Standard Deviation**: The given standard deviation of $20 shows the variability or spread of spending amounts around the mean.

These parameters allow businesses or researchers to predict the probability of encountering certain spending behaviors, crucial for understanding consumer habits.

Independent Events

When dealing with multiple events, such as predicting the spending habits of two different shoppers, the concept of independent events plays a crucial role.

Independent events mean that the occurrence of one event does not affect the other. In our context, if the spending of one shopper is independent of another, calculating the combined probability involves multiplying the probabilities of each event.

For example, to determine the probability of both shoppers spending more than $115, we calculate the probability for one shopper first. Since these events are independent, we then square this probability to find the combined likelihood of both shoppers exceeding this spending threshold.

Understanding these concepts helps in correctly analyzing joint probabilities, ensuring precise conclusions about consumer behavior. This is particularly important when considering random samples in statistics, where independent assumption allows for simplifying complex probability calculations.