Normal Distribution

Assume a character in a game is programmed to have an attack power defined by:
\(X \sim \text{Normal}(\mu = 40, \sigma = 3)\)

(a) Probability that the attack is greater than 45

Calculate \(\Pr(X > 45)\)

(b) Probability that the attack is between 32 and 42

Calculate \(\Pr(32 < X < 42)\)

(c) Probability that the sum of two attacks is greater than 85 units

Let \(X_1\) and \(X_2\) be the first and second attacks.
Find \(\Pr(X_1 + X_2 > 85)\)

(d) 99% limits for the sum of two attacks

Determine the values that define the central 99% range for \(X_1 + X_2\)

(e) Probability that the difference in attacks exceeds 5 units

Let \(D = X_1 - X_2\)
Calculate \(\Pr(|D| > 5) = \Pr(D < -5) + \Pr(D > 5)\) ```

Solutions

Here’s a step-by-step breakdown of the normal distribution questions using \(X \sim N(40, 3^2)\), i.e., a mean of 40 and standard deviation of 3:

(a) Probability that attack power > 45

\[ Z = \frac{45 - 40}{3} = \frac{5}{3} ≈ 1.667 \]

\[ P(X > 45) = P(Z > 1.667) ≈ 1 - 0.9525 = 0.0475 \]

Answer: Approximately 4.75%

(b) Probability that attack is between 32 and 42

Convert both to Z-scores:

  • \(Z_1 = \frac{32 - 40}{3} = -2.667\)
  • \(Z_2 = \frac{42 - 40}{3} = 0.667\)

\[ P(32 < X < 42) = P(-2.667 < Z < 0.667) = P(Z < 0.667) - P(Z < -2.667) ≈ 0.7475 - 0.0038 = 0.7437 \]

Answer: Approximately 74.37%

(c) Probability that the sum of two attacks is greater than 85

Sum of two independent normal variables:

\[ X_1 + X_2 \sim N(40+40,\ \sqrt{3^2+3^2}) = N(80,\ \sqrt{18}) = N(80,\ 4.243) \]

\[ Z = \frac{85 - 80}{4.243} ≈ 1.178 \] \[ P(X_1 + X_2 > 85) = P(Z > 1.178) ≈ 1 - 0.8801 = 0.1199 \]

Answer: Approximately 11.99%

(d) 99% limits for the sum of two attacks

Use the normal range formula:

\[ \text{Sum} \sim N(80,\ 4.243) \Rightarrow \text{99% interval} = \mu \pm z_{0.005} \cdot \sigma \]

Where \(z_{0.005} ≈ 2.576\)

\[ \text{Limits} = 80 \pm (2.576 \cdot 4.243) ≈ 80 \pm 10.93 \Rightarrow [69.07,\ 90.93] \]

Answer: Between 69.07 and 90.93 units

(e) Probability that the difference in attacks exceeds 5 units

Let \(D = X_1 - X_2\), where:

  • Mean \(\mu_D = 0\)
  • Variance \(\sigma_D^2 = 3^2 + 3^2 = 18\) ⇒ SD = 4.243

We compute:

\[ P(|D| > 5) = P(D < -5) + P(D > 5) \]

Calculate both tails:

\[ Z = \frac{5}{4.243} ≈ 1.178 \Rightarrow P(Z > 1.178) ≈ 0.1199 \]

So:

\[ P(|D| > 5) = 2 \cdot 0.1199 = 0.2398 \]

Answer: Approximately 23.98%