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Chapter 6: Problem 41

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditures for all customers at this supermarket will be normally distributed with a mean of $$\$ 95$$ and a standard deviation of $$\$ 20.$$ If the management wants to give free gifts to at most \(10 \%\) of the customers, what should the amount be above which a customer would receive a free gift?

Short Answer

Expert verified
The amount above which a customer would receive a free gift should be approximately \$121.60.

Step by step solution

01

Identify given data

The mean expenditure is given as \$95 and the standard deviation is \$20. Moreover, the management wants the top 10% of the customers (in terms of spending) to receive free gifts. This means we are interested in identifying the cutoff point above which these top 10% of customers fall.
02

Find the Z-score corresponding to the top 10%

Since we want to find the amount that 10% of customers exceed, we're actually interested in the cutoff for the bottom 90% (since 100% - 10% = 90%). We can find the Z-score for this in a standard normal distribution table or using a calculator, which is approximately 1.28.
03

Apply the Z-score to convert the normal distribution to standard normal distribution

In order to convert the problem into a value in our specific context (customer expenditures), we use the formula for converting a raw score (X) into a Z-score which is Z = (X - 渭) / 蟽, where 渭 is the mean and 蟽 is the standard deviation. Solving for X gives X = Z*蟽 + 渭.
04

Compute the spending limit for the top 10%

Substituting the Z-score, standard deviation, and mean values into this equation: X = 1.28 * 20 + 95 gives X 鈮� \$121.60. Therefore, the amount above which a customer would receive a gift should be set at \$121.60.

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Key Concepts

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

Mean
In any data set, the mean is the average value. It's found by adding up all the values and then dividing by the number of values.

In our context, the mean customer expenditure at the supermarket is \( \\(95 \). This means if we sum up all the spending and divide by the number of customers, we get \( \\)95 \).

This value is central, showing the typical spend amount, and is crucial for understanding customer behavior. Knowing the mean helps supermarkets set policies that cater to average spending.
  • Assumes data is distributed evenly around it.
  • Useful for calculating probabilities.
Standard Deviation
Standard deviation measures spread or variability in your data. It's the average amount by which data points differ from the mean.

In our problem, the standard deviation is \( \\(20 \), indicating customer expenditures usually vary by about \( \\)20 \) from the average spend of \( \$95 \).

This helps in understanding just how diverse spending habits are.

A smaller standard deviation means spending is more consistent, whereas a larger one means there's wider variance.
  • Gives insight into spending habit variety.
  • Essential for calculating the Z-score.
Z-score
A Z-score tells us how many standard deviations a data point is from the mean. It's a way to standardize values on a normal distribution, making it easier to calculate probabilities.

For our task, we need to find the Z-score for the top 10% of spenders, which corresponds to the bottom 90%, because 100% minus 10% equals 90%.

This Z-score is 1.28.

It's used to identify the cutoff point; the higher a Z-score, the further from the mean. The calculation is simplified with the formula \( Z = \frac{X - \mu}{\sigma} \), where \( X \) is the raw score.

This formula helps convert specific spending amounts into standardized values, vital for determining the spending limit for gifts.
  • Translates normal data into standard terms.
  • Helps find specific cutoff points.
Probability
Probability tells us how likely an event is to happen. In a normal distribution, it often relates to how data is spread across the range.

For the supermarket case, we want to know the probability that a customer spends above a certain amount, in this case, at the top 10% level. By setting the spending limit at approximately \( \$121.60 \), the management ensures only 10% of customers are likely to get a free gift.

This use of probability helps balance promotional costs with customer engagement.
  • Essential for making decisions and predictions.
  • Directly related to areas under the normal curve.

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Most popular questions from this chapter

For a binomial probability distribution, \(n=120\) and \(p=.60\). Let \(x\) be the number of successes in 120 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \leq 69)\) using the normal approximation. c. Find \(P(67 \leq x \leq 73)\) using the normal approximation.

For the standard normal distribution, find the area within one standard deviation of the mean - that is, the area between \(\mu-\sigma\) and \(\mu+\sigma .\)

Let \(x\) denote the time taken to run a road race. Suppose \(x\) is approximately normally distributed with a mean of 190 minutes and a standard deviation of 21 minutes. If one runner is selected at random, what is the probability that this runner will complete this road race a. in less than 160 minutes? b. in 215 to 245 minutes?

A machine at Keats Corporation fills 64 -ounce detergent jugs. The machine can be adjusted to pour, on average, any amount of detergent into these jugs. However, the machine does not pour exactly the same amount of detergent into each jug; it varies from jug to jug. It is known that the net amount of detergent poured into each jug has a normal distribution with a standard deviation of \(.35\) ounce. The quality control inspector wants to adjust the machine such that at least \(95 \%\) of the jugs have more than 64 ounces of detergent. What should the mean amount of detergent poured by this machine into these jugs be?

The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of $$\$ 845$$ and a standard deviation of $$\$ 270.$$ Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between $$\$ 1000$$ and $$\$ 1440? $$ b. What percentage of the Visa cards issued by this bank have a balance of $$\$ 730$$ or more?
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