Question 1192344
**1. Calculate the Mean and Standard Deviation of the Total Weight**

* **Mean:**
    * Mean weight of a single passenger = 189.8 pounds
    * Mean weight of 21 passengers = 189.8 pounds/passenger * 21 passengers = 3985.8 pounds

* **Standard Deviation:**
    * Standard deviation of a single passenger = 44.5 pounds
    * Standard deviation of the total weight of 21 passengers = 44.5 pounds/passenger * √21 passengers ≈ 204.4 pounds

**2. Standardize the Total Weight Limit**

* We'll use the z-score to standardize the total weight limit:

   z = (Total Weight Limit - Mean Total Weight) / Standard Deviation of Total Weight
   z = (4375 pounds - 3985.8 pounds) / 204.4 pounds
   z ≈ 1.90

**3. Find the Probability**

* We're interested in the probability that the total weight exceeds 4375 pounds, which corresponds to the area to the right of z = 1.90 under the standard normal distribution curve.

* Using a standard normal distribution table or a calculator, we find:
   P(Z > 1.90) ≈ 0.0287

**Therefore, the approximate probability that the total weight of the 21 passengers exceeds 4375 pounds is 0.029.**

**Note:**

* We've used the normal distribution as an approximation, even though passenger weights might not be perfectly normally distributed. 
* This approximation should be reasonably accurate due to the Central Limit Theorem, which states that the distribution of the sum or average of a large number of independent random variables tends to be approximately normal, regardless of the underlying distribution of the individual variables.