Q.No. 24 A normal random variable has mean equal to 0, and standard deviation equal to 3. The probability that on a random draw the value of this random variable is greater than 0 is __________ (round off to 2 decimal places).
The probability is 0.50. For a normal distribution symmetric about its mean of 0, exactly half the probability mass lies above the mean. The standard deviation of 3 does not affect this symmetry property.
Detailed Solution
A normal random variable X ∼ N(0,3) has a bell-shaped probability density function centered at μ = 0. Due to symmetry, P(X < 0) = P(X > 0) = 0.5. Standardizing to the standard normal Z = (X – 0)/3 gives P(X > 0) = P(Z > 0) = 1 – Φ(0) = 1 – 0.5 = 0.5, where Φ is the standard normal CDF.
No standardization is needed here since the threshold equals the mean. Rounded to two decimal places, the answer is 0.50.
Core Concept
In a N(0, σ²) distribution, symmetry ensures P(X > 0) = 0.5 for any σ > 0. Here, σ = 3 scales the spread but preserves the 50-50 split around zero.
Step-by-Step Calculation
- Identify X ∼ N(0,9) (variance = σ² = 9).
- Note P(X > 0) targets the mean.
- By definition, P(X > μ) = 0.5.
- No Z-table lookup required: directly 0.50 (rounded to 2 decimals).
Common Misconceptions
Students sometimes standardize unnecessarily: Z = X/3, so P(X > 0) = P(Z > 0) = 0.5. Others confuse with empirical rule (68-95-99.7%), which applies to intervals away from mean, not at the mean.
