Q.59 A new game is being introduced in a casino. A player can lose Rs. 100, break even, win Rs. 100, or win Rs. 500.
The probabilities (P(X)) of each of these outcomes (X) are given in the following table:
| X (in Rs.) | -100 | 0 | 100 | 500 |
|---|---|---|---|---|
| P(X) | 0.25 | 0.5 | 0.2 | 0.05 |
The standard deviation (σ) for the casino payout is Rs. ________ (rounded off to the nearest integer).
Given Casino Game Data
| Outcome X (Rs.) | -100 | 0 | 100 | 500 |
|---|---|---|---|---|
| Probability P(X) | 0.25 | 0.5 | 0.2 | 0.05 |
Step-by-Step Calculation
1. Expected Value (Mean)
μ = (-100 × 0.25) + (0 × 0.5) + (100 × 0.2) + (500 × 0.05)
= -25 + 0 + 20 + 25 = 20 Rs.
2. Variance
Compute (X − μ)² × P(X):
- (-100 − 20)² × 0.25 = 14400 × 0.25 = 3600
- (0 − 20)² × 0.5 = 400 × 0.5 = 200
- (100 − 20)² × 0.2 = 6400 × 0.2 = 1280
- (500 − 20)² × 0.05 = 230400 × 0.05 = 11520
σ² = 3600 + 200 + 1280 + 11520 = 16600
3. Standard Deviation
σ = √16600 ≈ 173 Rs. (rounded)
Final Answer
Standard deviation of casino payout = 173 Rs.
Interpretation
- High SD shows game outcomes vary widely
- Mean = +20 Rs indicates long-term positive expected return
- Volatility is significant despite positive expected value
Exam Tip
Always compute in order:
1) Mean → 2) (X−μ)² × P(X) → 3) Sum to get variance → 4) Square root → SD.
