SOLUTION: The amount of tries golf player 1 needs to score 18 holes is normally distributed with a mean of 92 and a deviation of 8. The amount of tries for golf player 2 to score 18 holes i

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Question 1199561: The amount of tries golf player 1 needs to score 18 holes is normally distributed with a mean of 92 and a deviation of 8.
The amount of tries for golf player 2 to score 18 holes is also normally distributed with a mean of 84 and a deviation of 9.
What is the probability that golf player 2 will lose against golf player 1?
Answer by ElectricPavlov(122) (Show Source):
You can put this solution on YOUR website!
**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%.**