Question 1202958
**1. Define Variables**

* Let X_i represent the weight of the i-th suitcase.
* Let S_n represent the sum of the weights of n suitcases: S_n = X_1 + X_2 + ... + X_n

**2. Determine the Mean and Standard Deviation of the Total Weight**

* **Mean of S_n:** 
    * E(S_n) = E(X_1) + E(X_2) + ... + E(X_n) = n * E(X_i) 
    * E(S_n) = n * 18 kg 

* **Standard Deviation of S_n:** 
    * Var(S_n) = Var(X_1) + Var(X_2) + ... + Var(X_n) (assuming independence of suitcase weights)
    * Var(S_n) = n * Var(X_i) = n * (5 kg)² = 25n 
    * Standard Deviation of S_n: σ_S_n = √(Var(S_n)) = √(25n) = 5√n

**3. Calculate the Probability of Exceeding the Weight Limit**

* We need to find P(S_n > 3500 kg) for each value of n.

* **Standardize the Total Weight:**
    * Z = (S_n - E(S_n)) / σ_S_n 
    * Z = (S_n - 18n) / (5√n)

* **Find the Critical Value (z_c):**
    * The weight limit is exceeded when S_n > 3500 kg.
    * We need to find the z-score (z_c) corresponding to the probability of exceeding the weight limit.

* **Calculate the Probability:**
    * P(S_n > 3500) = P(Z > z_c) 

**4. Calculations for Different Numbers of Suitcases**

* **(a) n = 190:**
    * E(S_190) = 190 * 18 = 3420 kg
    * σ_S_190 = 5√190 ≈ 68.56 kg
    * z_c = (3500 - 3420) / 68.56 ≈ 1.16
    * P(S_190 > 3500) = P(Z > 1.16) 
        * Use a standard normal distribution table to find this probability.

* **(b) n = 200:**
    * E(S_200) = 200 * 18 = 3600 kg
    * σ_S_200 = 5√200 ≈ 70.71 kg
    * z_c = (3500 - 3600) / 70.71 ≈ -1.41
    * P(S_200 > 3500) = P(Z > -1.41) = 1 - P(Z ≤ -1.41)

* **(c) n = 202:**
    * E(S_202) = 202 * 18 = 3636 kg
    * σ_S_202 = 5√202 ≈ 71.13 kg
    * z_c = (3500 - 3636) / 71.13 ≈ -1.91
    * P(S_202 > 3500) = P(Z > -1.91) = 1 - P(Z ≤ -1.91)

* **(d) n = 204:**
    * E(S_204) = 204 * 18 = 3672 kg
    * σ_S_204 = 5√204 ≈ 71.41 kg
    * z_c = (3500 - 3672) / 71.41 ≈ -2.41
    * P(S_204 > 3500) = P(Z > -2.41) = 1 - P(Z ≤ -2.41)

* **(e) n = 205:**
    * E(S_205) = 205 * 18 = 3690 kg
    * σ_S_205 = 5√205 ≈ 71.49 kg
    * z_c = (3500 - 3690) / 71.49 ≈ -2.66
    * P(S_205 > 3500) = P(Z > -2.66) = 1 - P(Z ≤ -2.66)

**Use a standard normal distribution table or statistical software to find the probabilities associated with the calculated z-scores for each case.**

**Important Notes:**

* This analysis assumes that the weights of the suitcases are independent and identically distributed.
* The Central Limit Theorem suggests that the distribution of the total weight (S_n) will tend towards a normal distribution as the number of suitcases (n) increases, even though the individual suitcase weights are not normally distributed.
* This calculation provides an approximation of the probability. 
* In reality, the weight distribution might not perfectly follow a normal distribution, which could affect the accuracy of the results.

I hope this comprehensive explanation helps! Let me know if you have any further questions.