Question 1192227
Here's how to solve this problem:

**i. Packets less than 2 kilograms (2000 grams):**

1. **Calculate the z-score:**
   z = (x - μ) / σ
   z = (2000 - 2025) / 15
   z = -1.67

2. **Find the probability:**
   Using a z-table or calculator, the probability of a z-score less than -1.67 is approximately 0.0475.

3. **Percentage:**
   0.0475 * 100% = 4.75%

**ii. Packets more than 2030 grams:**

1. **Calculate the z-score:**
   z = (2030 - 2025) / 15
   z = 0.33

2. **Find the probability:**
   The probability of a z-score greater than 0.33 is 1 - P(z < 0.33).  Using a z-table, P(z < 0.33) is approximately 0.6293. So, the probability of a z-score greater than 0.33 is 1 - 0.6293 = 0.3707.

3. **Percentage:**
   0.3707 * 100% = 37.07%

**iii. Adjusting the mean for 10% underweight:**

1. **Find the z-score for the 10th percentile:**
   Using a z-table or calculator, the z-score corresponding to 10% (0.10) is approximately -1.28.

2. **Use the z-score formula to find the new mean (μ_new):**
   x = μ_new + zσ
   2000 = μ_new + (-1.28)(15)
   2000 = μ_new - 19.2
   μ_new = 2000 + 19.2
   μ_new = 2019.2 grams

**iv. Adjusting the standard deviation for 10% less than 1995 grams:**

1. **Find the z-score for the 10th percentile:**
   As before, the z-score for 10% is approximately -1.28.

2. **Use the z-score formula to find the new standard deviation (σ_new):**
   x = μ + zσ_new
   1995 = 2025 + (-1.28)σ_new
   -30 = -1.28σ_new
   σ_new = 30 / 1.28
   σ_new ≈ 23.44 grams

**Summary of Answers:**

* i. Approximately 4.75% of packets are less than 2 kg.
* ii. Approximately 37.07% of packets are more than 2030 grams.
* iii. The manufacturer should adjust the mean to approximately 2019.2 grams.
* iv. The manufacturer should adjust the standard deviation to approximately 23.44 grams.