Probability of Grocery Expenses for Customers Paying by Check

Question:

What are the probabilities and percentages associated with grocery expenses for customers who pay by personal checks at supermarket grocery stores?

Answer:

To solve these probability questions, we will use the properties of the normal distribution and the provided mean and standard deviation. Let's calculate the probabilities for each scenario:

(a) Find the probability that a randomly selected customer who pays by check spends more than $114 on groceries.

To solve this, we need to calculate the area under the normal distribution curve to the right of $114. We'll use the z-score formula:

z = (x - μ) / σ

where

x = $114 (value we want to find the probability for)

μ = $87 (mean)

σ = $22 (standard deviation)

z = (114 - 87) / 22

z = 1.2273

Using a standard normal distribution table or a calculator, we can find the area to the right of the z-score of 1.2273, which represents the probability that a customer spends more than $114. Let's denote this probability as P(Z > 1.2273).

P(Z > 1.2273) ≈ 1 - P(Z < 1.2273)

Looking up the value in the table, we find that P(Z < 1.2273) ≈ 0.8898.

Therefore, P(Z > 1.2273) ≈ 1 - 0.8898 = 0.1102.

So, the probability that a randomly selected customer who pays by check spends more than $114 on groceries is approximately 0.1102 or 11.02%.

(b) What percentage of customers paying by check spend between $40 and $60 on groceries?

To solve this, we need to calculate the area under the normal distribution curve between $40 and $60. We'll convert these values to z-scores and then find the area between these z-scores.

(c) What percentage of customers paying by check spend between $70 and $105 on groceries?

We'll follow a similar approach as in part (b).

(d) Is it possible for a customer paying by check to spend more than $185? Explain.

To answer this question, we need to determine if the z-score corresponding to $185 is within the range of the distribution.

z = (185 - 87) / 22

z = 4.4545

Since the z-score of 4.4545 is quite high, it indicates that a customer spending $185 on groceries would be an extreme outlier in the distribution.

It's highly unlikely for a customer paying by check to spend more than $185, given the provided mean of $87 and standard deviation of $22. Therefore, it is highly improbable for a customer to spend more than $185 on groceries.