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The amount of money spent by a customer at a discount store has a mean of $$\$ 100$$ and a standard deviation of $$\$ 30.$$ What is the probability that a randomly selected group of 50 shoppers will spend a total of more than $$\$ 5300 ?$$ (Hint: The total will be more than \(\$ 5300\) when the sample average exceeds what value?)

Short Answer

Expert verified
The short answer is the final probability you obtained from Step 5 by calculating \(P(Z > calculated \, z-score)\).

Step by step solution

01

Analyse The Given Data

Here, the mean (\(μ\)) of the amount of money spent is $100 and the standard deviation (\(σ\)) is $30. The sample size (\(n\)) is 50 shoppers and the total amount in question is $5300. You're asked to find the probability that the total spend exceeds this amount.
02

Calculate The Sample Mean

The sample mean (\(X̄\)) is the given total amount ($5300) divided by the number of people (50). So, calculate \(X̄ = 5300/50\).
03

Find The Standard Error

Standard error is calculated by dividing the standard deviation by the square root of the sample size. Here, the standard deviation is $30 and the sample size is 50. The formula for standard error is \(σ/√n\), so calculate \(Standard \, Error = σ/√n = 30/√50\).
04

Calculate The Z-Score

The formula for z-score is \(Z = (X̄ - μ)/Standard \, Error\). From steps 2 & 3, we already know the sample mean and the standard error. So substitute these values into this formula to find the z-score.
05

Find The Probability

Use a z-table, calculator, or other resource to determine the probability associated with the calculated z-score. Additionally, since we want the probability of total spending exceeding $5300, it corresponds to the area under the standard normal curve to the right of the calculated z-value. The required probability is given by \(P(Z > calculated \, z-score)\). This gives you the final probability.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Deviation
Understanding the standard deviation in statistics is crucial for interpreting the spread of data. The standard deviation, symbolized as \( \sigma \), quantifies how much the values in a dataset deviate from the mean, or average. In simpler terms, it reflects how spread out the numbers are. A small standard deviation indicates that the data points tend to be close to the mean, while a large standard deviation suggests that the data are more spread out.

For example, in a retail context, if customer expenditures at a store have a low standard deviation, this implies that most customers spend roughly the same amount. But if that standard deviation is high, there is a wider variety of spending behaviors among customers. This kind of insight can be very valuable for a retail business when planning inventory, marketing strategies, and customer engagement.
Sample Mean
The sample mean, often denoted as \( \bar{X} \), is the average value of a sample and is used as an estimate of the population mean. To calculate the sample mean, you simply sum up all the values in the sample and then divide by the number of observations. This can be particularly useful in instances where collecting data for the entire population is impractical or impossible.

Using the retail example from the exercise, by calculating the average amount spent by a group of 50 shoppers, we can estimate what a typical shopper might spend. This helps in understanding customer behavior on average, rather than focusing on any one individual's spending, which could be anomalously high or low.
Standard Error
The standard error measures the accuracy with which a sample mean represents the population mean. Mathematically, it's the standard deviation of the sampling distribution of the sample mean, denoted by \( SE \). The formula for finding the standard error is \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the standard deviation of the sample and \( n \) is the number of observations in the sample.

In practical terms, a smaller standard error means the sample mean is closer to the true population mean. This is an essential concept when determining how confident we can be in our sample estimates. In the textbook problem, calculating the standard error allows us to judge the reliability of the sample mean of the customer expenditures we've calculated.
Z-Score
A z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. The z-score indicates how many standard deviations an element is from the mean. It's calculated using the formula \( Z = \frac{(X - \mu)}{SE} \), where \( X \) is the value, \( \mu \) is the mean, and \( SE \) is the standard error.

In the context of our exercise, the z-score is used to find the probability that the shoppers will spend more than a certain amount. It's a powerful tool because it enables comparison across different sets of data which may have different means and standard deviations. After finding the z-score, we can refer to a z-table to find the probability of a value occurring within a normal distribution, which is fundamental in making predictions based on statistical data.

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Most popular questions from this chapter

The nicotine content in a single cigarette of a particular brand has a distribution with mean \(0.8 \mathrm{mg}\) and standard deviation \(0.1 \mathrm{mg}\). If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than \(0.79 ?\) less than \(0.77\) ?

Suppose that \(20 \%\) of the subscribers of a cable television company watch the shopping channel at least once a week. The cable company is trying to decide whether to replace this channel with a new local station. A survey of 100 subscribers will be undertaken. The cable company has decided to keep the shopping channel if the sample proportion is greater than \(.25 .\) What is the approximate probability that the cable company will keep the shopping channel, even though the true proportion who watch it is only \(.20 ?\)

A certain chromosome defect occurs in only 1 out of 200 adult Caucasian males. A random sample of \(n=100\) adult Caucasian males is to be obtained. a. What is the mean value of the sample proportion \(p\), and what is the standard deviation of the sample proportion? b. Does \(p\) have approximately a normal distribution in this case? Explain. c. What is the smallest value of \(n\) for which the sampling distribution of \(p\) is approximately normal?

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within \(0.5\) of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7\) ?

Newsweek (November 23, 1992) reported that 40\% of all U.S. employees participate in "self-insurance" health plans \((\pi=.40)\). a. In a random sample of 100 employees, what is the approximate probability that at least half of those in the sample participate in such a plan? b. Suppose you were told that at least 60 of the \(100 \mathrm{em}\) ployees in a sample from your state participated in such a plan. Would you think \(\pi=.40\) for your state? Explain.

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