Question

A six-face fair die is rolled once, with X being the number that appeared on the uppermost surface. Then the variance of X is ________ (rounded off to three decimal places).

Answer 1

Correct Answer: 2.9

To calculate the variance of \(X\), the number that appears on the uppermost surface of a six-faced fair die, we start by identifying the possible outcomes and their probabilities. Since the die is fair, each face has an equal probability of appearing when rolled.

Possible Outcomes: 1, 2, 3, 4, 5, 6 

Probability of Each Outcome: \(\frac{1}{6}\)

Step 1: Calculate Expected Value (Mean) \(\mu\)
\(\mu = E(X) = \sum (x \cdot P(x))\)
\(\mu = (1 \times \frac{1}{6}) + (2 \times \frac{1}{6}) + (3 \times \frac{1}{6}) + (4 \times \frac{1}{6}) + (5 \times \frac{1}{6}) + (6 \times \frac{1}{6})\)
\(\mu = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5\)

Step 2: Calculate Variance \(\sigma^2\)
\(\sigma^2 = E(X^2) - (E(X))^2\)
\(E(X^2) = (1^2 \times \frac{1}{6}) + (2^2 \times \frac{1}{6}) + (3^2 \times \frac{1}{6}) + (4^2 \times \frac{1}{6}) + (5^2 \times \frac{1}{6}) + (6^2 \times \frac{1}{6})\)
\(E(X^2) = (1 + 4 + 9 + 16 + 25 + 36) \times \frac{1}{6} = \frac{91}{6}\)
\(\sigma^2 = \frac{91}{6} - (3.5)^2\)
\(\sigma^2 = \frac{91}{6} - 12.25\)
\(\sigma^2 = 15.1667 - 12.25\)
\(\sigma^2 = 2.9167\)

The variance of \(X\) is 2.917 (rounded to three decimal places), which fits within the expected range (2.9, 2.9).