Question 1198453: Armstrong Faber produces a standard number-two pencil called Ultra-Lite. Since Chuck Armstrong started Armstrong Faber, sales have grown steadily. With the Increase in the price of wood products. however, Chuck has been forced to increase the price of the Ultra-Lite pencils. As a result, the demand for Ultra-Lite has been fairly stable over the past 6 years. On the average, Armstrong Faber has sold 457,000 pencils each year. Furthermore, 90% of the time sales have been between 454,000 and 460,000 pencils. It is expected that the sales follow a normal distribution with a mean of 457,000 pencils. Estimate the standard deviation of this distribution.
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! **1. Define**
* **μ (mu):** Mean sales = 457,000 pencils
* **σ (sigma):** Standard deviation (which we need to find)
* **X1:** Lower limit of sales = 454,000 pencils
* **X2:** Upper limit of sales = 460,000 pencils
**2. Standardize the Values**
* We need to convert the sales values (X1 and X2) to z-scores:
* z = (X - μ) / σ
* z1 = (X1 - μ) / σ = (454,000 - 457,000) / σ = -3,000 / σ
* z2 = (X2 - μ) / σ = (460,000 - 457,000) / σ = 3,000 / σ
**3. Find the Corresponding Z-scores for 90% of the Data**
* Since 90% of the sales fall between 454,000 and 460,000 pencils, we need to find the z-scores that correspond to the middle 90% of the data in a standard normal distribution.
* This means 5% of the data will be below the lower z-score (z1), and 5% will be above the upper z-score (z2).
* Using a standard normal distribution table or a calculator:
* The z-score corresponding to the 5th percentile is approximately -1.645.
* The z-score corresponding to the 95th percentile is approximately 1.645.
**4. Set Up and Solve the Equations**
* We know:
* z1 = -1.645 = -3,000 / σ
* z2 = 1.645 = 3,000 / σ
* **Solve for σ:**
* From either equation: σ = 3,000 / 1.645
* σ ≈ 1820.3
**Therefore, the estimated standard deviation of the sales distribution is approximately 1,820 pencils.**
**In summary:**
* We used the given information about the mean, the range of sales for 90% of the data, and the properties of the normal distribution to estimate the standard deviation of the sales.
* This estimate helps understand the variability in sales and can be useful for forecasting and inventory management.
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