Question 1200047
**1. Check for Normal Approximation**

* **n = 138 (sample size)**
* **p = 0.37 (probability of success - claim padded)**
* **q = 1 - p = 0.63 (probability of failure - claim not padded)**

* **np = 138 * 0.37 = 51.06 ≥ 5** 
* **nq = 138 * 0.63 = 86.94 ≥ 5**

Since both np and nq are greater than 5, the normal approximation to the binomial distribution is appropriate.

**2. Calculate Mean (μ) and Standard Deviation (σ)**

* **μ = np = 138 * 0.37 = 51.06**
* **σ = √(npq) = √(138 * 0.37 * 0.63) = 5.67**

**3. Standardize and Use Standard Normal Table (or Calculator)**

* **(a) Half or more of the claims have been padded:**
    * P(X ≥ 69) 
    * **Continuity Correction:** P(X ≥ 68.5) 
    * **Standardize:** z = (68.5 - 51.06) / 5.67 = 3.09 
    * **P(Z ≥ 3.09) = 0.0010** (using a standard normal table or calculator)

* **(b) Fewer than 45 of the claims have been padded:**
    * P(X < 45) 
    * **Continuity Correction:** P(X < 44.5) 
    * **Standardize:** z = (44.5 - 51.06) / 5.67 = -1.17 
    * **P(Z < -1.17) = 0.1210**

* **(c) From 40 to 64 of the claims have been padded:**
    * P(40 ≤ X ≤ 64) 
    * **Continuity Correction:** P(39.5 ≤ X ≤ 64.5) 
    * **Standardize:** 
        * z1 = (39.5 - 51.06) / 5.67 = -2.07 
        * z2 = (64.5 - 51.06) / 5.67 = 2.39 
    * **P(-2.07 ≤ Z ≤ 2.39) = 0.9817 - 0.0192 = 0.9625**

* **(d) More than 80 of the claims have not been padded:**
    * If 80 claims have not been padded, then 58 claims have been padded.
    * P(X > 58) 
    * **Continuity Correction:** P(X > 58.5) 
    * **Standardize:** z = (58.5 - 51.06) / 5.67 = 1.31 
    * **P(Z > 1.31) = 0.0951** 

**Therefore:**

* **(a) Probability of half or more of the claims being padded: 0.0010**
* **(b) Probability of fewer than 45 claims being padded: 0.1210**
* **(c) Probability of from 40 to 64 claims being padded: 0.9625**
* **(d) Probability of more than 80 claims not being padded: 0.0951**