Let X denote the number of scores which exceed 4 in 18, tosses of a symmetrical die. Consider the following statements:
1. The arithmetic mean of X is 6.
2. The standard deviation of X is 2.
Which of the above statements is/are correct?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to analyze the situation where we are tossing a symmetrical die 18 times and counting the number of scores that exceed 4. The possible scores that exceed 4 on a die are 5 and 6.

### Step-by-Step Solution:

1. **Define the Random Variable**:
Let \( X \) be the number of scores that exceed 4 in 18 tosses of a symmetrical die.

2. **Identify the Number of Trials**:
The number of trials \( n \) is 18, as we are tossing the die 18 times.

3. **Determine the Probability of Success**:
The successful outcomes (scores exceeding 4) on a die are 5 and 6. Thus, the probability of success \( p \) is:
\[ p = \frac{\text{Number of successful outcomes}}{\text{Total outcomes}} = \frac{2}{6} = \frac{1}{3} \]

4. **Calculate the Probability of Failure**:
The probability of failure \( q \) is:
\[ q = 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \]

5. **Calculate the Arithmetic Mean**:
The arithmetic mean \( \mu \) of a binomial distribution is given by:
\[ \mu = n \cdot p = 18 \cdot \frac{1}{3} = 6 \]
Therefore, the first statement is correct.

6. **Calculate the Standard Deviation**:
The standard deviation \( \sigma \) of a binomial distribution is given by:
\[ \sigma = \sqrt{n \cdot p \cdot q} = \sqrt{18 \cdot \frac{1}{3} \cdot \frac{2}{3}} = \sqrt{18 \cdot \frac{2}{9}} = \sqrt{4} = 2 \]
Therefore, the second statement is also correct.

7. **Conclusion**:
Both statements are correct. Hence, the answer is that both statements 1 and 2 are correct.

### Final Answer:
Both statements are correct.

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