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A company that has a large number of supermarket grocery stores claims that customers who pay by personal checks spend an average of $$\$ 87$$ on groceries at these stores with a standard deviation of $$\$ 22 .$$ Assume that the expenses incurred on groceries by all such customers at these stores are normally distributed. a. Find the probability that a randomly selected customer who pays by check spends more than $$\$ 114$$ on groceries. b. What percentage of customers paying by check spend between $$\$ 40$$ and $$\$ 60$$ on groceries? c. What percentage of customers paying by check spend between $$\$ 70$$ and $$\$ 105 ?$$ d. Is it possible for a customer paying by check to spend more than $$\$ 185$$ ? Explain.

Short Answer

Expert verified
a. The calculation yields a Z-score of 1.23, which translates to a probability of approximately 10.9% for a customer to spend more than $114. \n b. The Z-scores for $40 and $60 are -2,13 and -1.23 respectively. According to the Z-table, this translates to approximately 1.33%.\n c. The Z-scores for $70 and $105 are -0.77 and 0.82 respectively. This yields a difference in probabilities and thus percentage of approximately 37%. \n d. However unlikely, it's statistically possible that a customer may spend more than $185.

Step by step solution

01

Understand the given data

This exercise provides a mean value (average value) of grocery spendings of \$87 per customer with a standard deviation of \$22. These values are applicable to all parts of the exercise. The data follows a normal distribution.
02

Calculating Z-Scores

The Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. It can be calculated by the formula, \(Z = \frac{(X - µ)}{σ}\) where X is the variable, µ is the mean and σ is the standard deviation. In part a, the Z-score can be calculated using the provided value of \$114. Apply the formula, \(Z = \frac{(114 - 87)}{22}\) that results in \(Z = 1.23\)
03

Find the probability for part a

This can be done by looking up this Z-score in a standard normal distribution table, or by using a calculator with a normal distribution function. The result will give the probability of a customer spending less than \$114, while the exercise asks for the probability of spending more than that. So, subtract this probability from 1 to get the solution.
04

Repeat Step 2 and 3 for parts b and c

For part b, calculate the Z-scores for both \$40 and \$60, then find the two corresponding probabilities and find the difference in these probabilities. This will be the required percentage. For part c, follow the same process using the values \$70 and \$105 instead.
05

Answer part d

As for part d, notably statistical possibilities don't entirely rule out the respective occurrence. A customer paying by check could still spend more than \$185, but the probability for this happening can be calculated and is expected to be very low. For this, calculate the Z-score for \$185 using the same formula and find the probability. The higher the value above the mean (\$87), the lower the probability.

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Key Concepts

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

Z-score
The Z-score is a nifty tool that helps us figure out how far a certain value is from the mean of a data set in terms of standard deviations. Imagine it as a ruler telling you where you stand in a crowd. The Z-score is calculated by the formula \( Z = \frac{(X - \mu)}{\sigma} \), where:
  • \( X \) is the particular value or observation you're interested in.
  • \( \mu \) is the mean of the data set.
  • \( \sigma \) is the standard deviation.
A Z-score of zero means the value is exactly the same as the mean.
Positive Z-scores indicate values above the mean, and negative ones tell you that you're below it.
In our example, calculating the Z-score for a spender using \( \$ 114 \) leads to a score of \( 1.23 \). This means the spending is 1.23 standard deviations above the average spending.
Probability
Probability is all about the chances of different outcomes. It helps answer questions like, "What's the likelihood of someone spending more than a certain amount?"
In statistics, we often use the concept of the normal distribution to find probabilities. The normal distribution is a bell-shaped curve that represents the distribution of data. Most values cluster around the mean, and the probabilities for extreme values decrease as you move away.
To find the probability related to a specific Z-score, we use the standard normal distribution table. This table gives us the probability that a value is below a given Z-score.
For instance, once the Z-score is calculated, you can check the table to know how much less than a particular score is likely, and adjust accordingly to find probabilities above that score. Understanding probability is like having a radar for anticipating patterns in data.
Standard Deviation
Standard deviation shows us how much variation or "spread" exists from the average (mean) value. It's a measure of how dispersed your data is. A small standard deviation means data points tend to be close to the mean. Conversely, a large standard deviation means they're spread out over a wider range of values.
In the context of spending at grocery stores, the standard deviation is \( \\( 22 \). This clues us in that while the mean spending is \( \\) 87 \), actual spending will differ, ranging around this average.
A key formula for standard deviation in a normal distribution is linked with the Z-score formula:
\( Z = \frac{(X - \mu)}{\sigma} \).
Recognizing standard deviation empowers you to grasp how varied the data points are compared to the mean.
Mean
The mean, or average, is the center point of a data set. If you think of a number line, the mean is typically right there in the middle where balance is maintained. It's calculated by summing all values and dividing by the count of those values.
In our grocery store example, the mean spending is \( \\( 87 \), meaning if you summed all customer spending and divided by the number of check-paying customers, you'd get \( \\) 87 \).
The mean is crucial because it gives you a central reference point, especially in normally distributed data.
It helps us not only identify the "typical" value but is also a key part of the puzzle when calculating Z-scores and assessing probabilities.
With the standard deviation, it helps paint a complete picture of the data's behavior and spread.

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Most popular questions from this chapter

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?

According to the records of an electric company serving the Boston area, the mean electricity consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electricity consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. a. Find the probability that the monthly electricity consumption during winter by a randomly selected household from this area is less than 1950 kilowatt- hours. b. What percentage of the households in this area have a monthly electricity consumption of 900 to 1300 kilowatt-hours?

How do the width and height of a normal distribution change when its mean remains the same but its standard deviation decreases?

Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

Let \(x\) be a continuous random variable. What is the probability that \(x\) assumes a single value, such as \(a\) ?

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