You intend to draw a random sample of size \( n=556 \) from a population whose parameter is \( p=0.83 \) What is the mean of the distribution of sample proportions? \( \mu_{\widehat{p}}=0.83 \) What is the standard deviation of the distribution of sample proportions? (Report answer accurate to 2 decimal places.) \( \sigma_{\widehat{p}}=\square \)

To find the standard deviation of the distribution of sample proportions, you can use the formula: \[ \sigma_{\widehat{p}} = \sqrt{\frac{p(1-p)}{n}} \] Plugging in the values \( p = 0.83 \) and \( n = 556 \): \[ \sigma_{\widehat{p}} = \sqrt{\frac{0.83(1-0.83)}{556}} = \sqrt{\frac{0.83 \times 0.17}{556}} \approx \sqrt{\frac{0.1411}{556}} \approx \sqrt{0.000253} \approx 0.016 \] Rounded to two decimal places, the standard deviation of the distribution of sample proportions \( \sigma_{\widehat{p}} \approx 0.02 \).