Normal Distribution: Arab Horses – Worked Example

The mass of Arab horses is assumed to follow a normal distribution with: - Mean (\(\mu\)) = 900 lbs
- Standard deviation (\(\sigma\)) = 50 lbs

Questions

  1. Calculate the probability that an Arab horse weighs more than 940 lbs.
  2. Calculate the probability that an Arab horse weighs between 880 lbs and 960 lbs.
  3. Find the weight exceeded by 97.5% of Arab horses.

Solution to Question 1: \(P(X > 940)\)

Let \(X\) be the mass of Arab horses.

  • Convert to Z-score: \[ Z = \frac{940 - 900}{50} = 0.8 \]
  • Use normal tables: \[ P(X \leq 940) = P(Z \leq 0.8) \approx 0.2119 \]
  • Therefore: \[ P(X > 940) = 1 - 0.2119 = \mathbf{0.7881} \]

Solution to Question 2: \(P(880 \leq X \leq 960)\)

We use the complement rule: \[ P(880 \leq X \leq 960) = 1 - [P(X \leq 880) + P(X \geq 960)] \]

Step 1: Compute each bound

  • Z-score for 960: \[ Z = \frac{960 - 900}{50} = 1.2 \quad \Rightarrow P(Z \geq 1.2) \approx 0.1151 \]
  • Z-score for 880: \[ Z = \frac{880 - 900}{50} = -0.4 \quad \Rightarrow P(Z \leq -0.4) = P(Z \geq 0.4) \approx 0.3446 \]

Step 2: Sum the tails

\[ P(\text{Outside Interval}) = 0.3446 + 0.1151 = 0.4597 \] \[ P(880 \leq X \leq 960) = 1 - 0.4597 = \mathbf{0.5403} \]

Solution to Question 3: Weight Exceeded by 97.5% of Horses

We want \(X_0\) such that: \[ P(X \geq X_0) = 0.975 \Rightarrow P(X \leq X_0) = 0.025 \]

  • Use standard normal table: \[ P(Z \leq -1.96) = 0.025 \]
  • Back-calculate \(X_0\): \[ -1.96 = \frac{X_0 - 900}{50} \Rightarrow X_0 = 900 + (-1.96 \cdot 50) = \mathbf{802} \text{ lbs} \]