Question

The variance of the number obtained in a throw of an unbiased die is:

• A. \(\frac{91}{6}\)
• B. \(\frac{35}{12}\)
• C. \(\frac{4}{3}\)
• D. \(\frac{21}{6}\)

Answer 1

The Correct Option is B

To find the variance of the number obtained in a throw of an unbiased die, we will follow these steps:

1. An unbiased die has 6 faces with numbers 1 to 6. Each number has an equal probability of appearing, which is \(\frac{1}{6}\).

2. The expected value \(E(X)\) of the numbers rolled can be calculated using the formula:

\(E(X) = \sum_{i=1}^{6} x_i \cdot P(x_i) = 1\cdot\frac{1}{6} + 2\cdot\frac{1}{6} + 3\cdot\frac{1}{6} + 4\cdot\frac{1}{6} + 5\cdot\frac{1}{6} + 6\cdot\frac{1}{6}\)

3. Calculating the above gives:

\(E(X) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5\)

4. The variance \(Var(X)\) is calculated using the formula:

\(Var(X) = E(X^2) - (E(X))^2\)

5. First, calculate \(E(X^2)\):

\(E(X^2) = \sum_{i=1}^{6} x_i^2 \cdot P(x_i) = 1^2\cdot\frac{1}{6} + 2^2\cdot\frac{1}{6} + 3^2\cdot\frac{1}{6} + 4^2\cdot\frac{1}{6} + 5^2\cdot\frac{1}{6} + 6^2\cdot\frac{1}{6}\)

6. Calculating the above gives:

\(E(X^2) = \frac{1+4+9+16+25+36}{6} = \frac{91}{6}\)

7. Substitute the values into the variance formula:

\(Var(X) = \frac{91}{6} - (3.5)^2\)

8. Calculate \((3.5)^2 = 12.25\), then:

\(Var(X) = \frac{91}{6} - \frac{12.25 \cdot 6}{6} = \frac{91}{6} - \frac{73.5}{6} = \frac{91 - 73.5}{6} = \frac{17.5}{6}\)

9. Simplifying \(\frac{17.5}{6}\):

\(\frac{17.5}{6} = \frac{35}{12}\)

Therefore, the correct variance is \(\frac{35}{12}\).