Correct Answer - Option 3 : 2500
Concept:
\(A = Z\sqrt{\frac{{P\left( {1 - P} \right)}}{n}}\)
\(A^2=Z^2\left[\frac{P(1-P)}{n}\right]\)
\(n=Z^2\left[\frac{P(1-P)}{A^2}\right]\)
where
Z = Standard normal variate whose value depends upon the confidence level
n = No. of observations
z = 2 (approx) (For 95% confidence level)
P = Percentage occurrence of an activity
A = Limit of accuracy percentage
Calculation:
Given:
Percentage occurrence of an activity (P) = 50 % = 0.5
Limit of accuracy (A) = \(\pm 2\% = 0.02\)
Confidence level = 95%
\(n=Z^2\left[\frac{P(1-P)}{A^2}\right]\)
\(n = \frac{{{{2}^2}\times 0.5\left( {1 - 0.5} \right)}}{{{{0.02}^{2\;}}}}=2500\)
