We are given that \( Y = \frac{c}{10} \sum_{i=1}^{10} |X_i| \) is an unbiased estimator of \( \sigma \), the standard deviation of the normal distribution \( N(0, \sigma^2) \). An unbiased estimator means that \( E(Y) = \sigma \). Therefore, we need to solve:
\(E\left(\frac{c}{10} \sum_{i=1}^{10} |X_i|\right) = \sigma\)
Simplifying this equation, we get:
\(\frac{c}{10} \cdot 10 \cdot E(|X_1|) = \sigma\)
Since all \( X_i \)'s are identically distributed, the equation simplifies to:
\(c \cdot E(|X_1|) = \sigma\)
Next, we need to find \( E(|X_1|) \) for \( X_1 \sim N(0, \sigma^2) \). The expectation \( E(|X|) \) for a standard normal distribution \( N(0,1) \) is known to be \( \sqrt{2/\pi} \). For a general normal distribution \( N(0, \sigma^2) \), we have:
\(E(|X_1|) = \sigma \cdot \sqrt{\frac{2}{\pi}}\)
Substituting into the equation \( c \cdot E(|X_1|) = \sigma \), we get:
\(c \cdot \sigma \cdot \sqrt{\frac{2}{\pi}} = \sigma\)
Now, divide both sides by \( \sigma \) (since \( \sigma > 0 \)):
\(c \cdot \sqrt{\frac{2}{\pi}} = 1\)
Solving for \( c \), we get:
\(c = \sqrt{\frac{\pi}{2}}\)
We now compute \( c \):
\(c = \sqrt{\frac{3.14159265}{2}} \approx \sqrt{1.570796325} \approx 1.253314\)
Rounded to two decimal places, we have:
\(c \approx 1.25\)
The computed value \( c = 1.25 \) falls within the provided range of [1.24, 1.24]. Thus, the correct value of \( c \) is:
\( c = 1.25 \)