To solve these problems, we will use properties of the normal distribution, specifically the probability density function (PDF) and the cumulative distribution function (CDF).

Given dataTo solve these problems, we will use properties of the normal distribution, specifically the probability density function (PDF) and the cumulative distribution function (CDF).

Given

• Mean demand, $\mu = 500$ bags
• Standard deviation, $\sigma = 75$ bags

The demand during the lead time period follows a normal distribution: $X \sim N(500, 75^2)$.

Solution

a) Probability that demand will be less than 400 bags:

To find $P(X < 400)$, we will standardize 400 to a Z-score:

$Z = \frac{X - \mu}{\sigma} = \frac{400 - 500}{75} = \frac{-100}{75} \approx -1.33$

Using the standard normal distribution table or a calculator, we find: $P(Z < -1.33) \approx 0.0918$

So, the probability that demand will be less than 400 bags is approximately 0.0918 (or 9.18%).

b) Probability that demand will be less than 550 bags:

To find $P(X < 550)$:

$Z = \frac{550 - 500}{75} = \frac{50}{75} \approx 0.67$

From the standard normal distribution: $P(Z < 0.67) \approx 0.7486$

The probability that demand will be less than 550 bags is approximately 0.7486 (or 74.86%).

c) Probability that demand will be equal to 500 bags:

In a continuous probability distribution, the probability of an exact value is zero. Therefore,

$P(X = 500) = 0$

d) Probability that demand will be between 550 and 600 bags:

To find $P(550 < X < 600)$, we calculate the Z-scores for 550 and 600, then find the area between these two values.

For $X = 550$: $Z = \frac{550 - 500}{75} \approx 0.67$

For $X = 600$: $Z = \frac{600 - 500}{75} \approx 1.33$

Using the standard normal distribution: $P(Z < 1.33) \approx 0.9082 \quad \text{and} \quad P(Z < 0.67) \approx 0.7486$

Thus, $P(550 < X < 600) = P(Z < 1.33) - P(Z < 0.67) \approx 0.9082 - 0.7486 = 0.1596$

The probability that demand will be between 550 and 600 bags is approximately 0.1596 (or 15.96%).

e) Reorder point for a 95% probability of meeting demand during lead time:

To ensure a 95% probability that demand does not exceed the inventory level, we need to find the demand level (reorder point) corresponding to the 95th percentile of the distribution.

For a standard normal distribution, the Z-score corresponding to the 95th percentile is approximately 1.645.

So, we set up the equation: $X = \mu + Z \cdot \sigma = 500 + 1.645 \cdot 75$

Calculating this: $X \approx 500 + 123.375 = 623.375$

Rounding up, the reorder point is approximately 624 bags.

Summary of Answers:

• a) Probability demand < 400 bags: 0.0918
• b) Probability demand < 550 bags: 0.7486
• c) Probability demand = 500 bags: 0
• d) Probability demand between 550 and 600 bags: 0.1596
• e) Reorder point for 95% probability: 624 bags

Would you like further details on any part of this solution or have any questions?

Here are five related questions to consider:

1. How does changing the lead time affect the reorder point for a 95% probability of meeting demand?
2. What would be the impact on probabilities if the standard deviation increased?
3. How would the reorder point change if the store wanted a 99% probability of meeting demand?
4. If the demand mean shifted to 550 bags, how would this affect each probability calculation?
5. How could you use this approach to set inventory levels for items with seasonal demand?

Tip: Understanding the role of standard deviation in inventory management is crucial, as it helps account for variability in demand, which impacts both reorder points and inventory levels.