The management of a supermarket want 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?

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**Title: Promotional Policy Analysis for a Supermarket**

**Introduction:**

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. 

**Problem Statement:**

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?

**Analysis:**

Given that the expenditures are normally distributed with:
- **Mean (µ)** = $95
- **Standard deviation (σ)** = $20

We need to find the expenditure amount (let's denote it as *X*) above which the top 10% of customers' expenditures lie. 

### Solution:

1. **Determine the Z-score** corresponding to the top 10%. The cumulative probability for the bottom 90% is 0.90. Using a Z-table or standard normal distribution calculator, the Z-score for 0.90 is approximately 1.28 (since 90% of the data lies below this point).

2. **Apply the Z-score formula** to find X:
\[ Z = \frac{X - \mu}{\sigma} \]

Where:
- \( Z \) = Z-score corresponding to 0.90 (top 10%)
- \( \mu \) = mean expenditure ($95)
- \( \sigma \) = standard deviation ($20)

\[ 1.28 = \frac{X - 95}{20} \]

3. **Solve for X**:
\[ X - 95 = 1.28 \times 20 \]
\[ X - 95 = 25.6 \]
\[ X = 95 + 25.6 \]
\[ X = 120.6 \]

Therefore, to ensure that only the top 10% of customers receive a free gift, the supermarket should set the expenditure threshold above $120.60.

---

This analysis can be utilized by educational institutions to demonstrate practical applications of normal distribution and Z-scores in promotional and business policy planning.
Transcribed Image Text:--- **Title: Promotional Policy Analysis for a Supermarket** **Introduction:** 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. **Problem Statement:** 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? **Analysis:** Given that the expenditures are normally distributed with: - **Mean (µ)** = $95 - **Standard deviation (σ)** = $20 We need to find the expenditure amount (let's denote it as *X*) above which the top 10% of customers' expenditures lie. ### Solution: 1. **Determine the Z-score** corresponding to the top 10%. The cumulative probability for the bottom 90% is 0.90. Using a Z-table or standard normal distribution calculator, the Z-score for 0.90 is approximately 1.28 (since 90% of the data lies below this point). 2. **Apply the Z-score formula** to find X: \[ Z = \frac{X - \mu}{\sigma} \] Where: - \( Z \) = Z-score corresponding to 0.90 (top 10%) - \( \mu \) = mean expenditure ($95) - \( \sigma \) = standard deviation ($20) \[ 1.28 = \frac{X - 95}{20} \] 3. **Solve for X**: \[ X - 95 = 1.28 \times 20 \] \[ X - 95 = 25.6 \] \[ X = 95 + 25.6 \] \[ X = 120.6 \] Therefore, to ensure that only the top 10% of customers receive a free gift, the supermarket should set the expenditure threshold above $120.60. --- This analysis can be utilized by educational institutions to demonstrate practical applications of normal distribution and Z-scores in promotional and business policy planning.
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