Question 1199561
**1. Define Variables**

* Let X1 be the number of tries for golf player 1. 
* Let X2 be the number of tries for golf player 2.

**2. Standardize the Variables**

* **Player 1:**
    * Z1 = (X1 - μ1) / σ1 
        * where:
            * μ1 = mean for player 1 (92)
            * σ1 = standard deviation for player 1 (8)

* **Player 2:**
    * Z2 = (X2 - μ2) / σ2 
        * where:
            * μ2 = mean for player 2 (84)
            * σ2 = standard deviation for player 2 (9)

**3. Determine the Condition for Player 2 to Lose**

* Player 2 loses if X2 > X1 
* This is equivalent to X2 - X1 > 0

**4. Find the Distribution of the Difference (X2 - X1)**

* The difference between two normally distributed variables is also normally distributed.
* Mean of (X2 - X1) = μ2 - μ1 = 84 - 92 = -8
* Variance of (X2 - X1) = σ1² + σ2² = 8² + 9² = 145
* Standard Deviation of (X2 - X1) = √145 ≈ 12.04

**5. Standardize the Difference**

* Let Z = (X2 - X1) 
* Z' = (Z - μ_Z) / σ_Z 
    * where:
        * μ_Z = -8 
        * σ_Z = 12.04

**6. Calculate the Probability of Player 2 Losing (Z > 0)**

* P(Player 2 Loses) = P(X2 - X1 > 0) 
* P(Player 2 Loses) = P(Z > (0 - (-8)) / 12.04) 
* P(Player 2 Loses) = P(Z > 0.66)

* Use a standard normal distribution table or calculator to find P(Z > 0.66) 
* P(Z > 0.66) ≈ 0.2546

**Therefore, the probability that golf player 2 will lose against golf player 1 is approximately 0.2546 or 25.46%.**