# Validation of Normal Approximation for Multiple Samples ## Introduction For each sample we verify whether the normal approximation to binomial distribution is appropriate based on the conditions: - \( np \geq 5 \) - \( n(1 - p) \geq 5 \) ## Sample Analyses ### Primary Sample (n = 65) - \( np = 65 \times .78 = 50.7 \geq 5 \) - \( n(1 - p) = 65 \times .22 = 14.3 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ### Secondary Sample (n = 52) - \( np = 52 \times .78 = 40.56 \geq 5 \) - \( n(1 - p) = 52 \times .22 = 11.44 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ### Auxiliary Sample (n = 71) - \( np = 71 \times .78 = 55.38 \geq 5 \) - \( n(1 - p) = 71 \times .22 = 15.62 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ### Ad Hoc Group (n = 48) - \( np = 48 \times .78 = 37.44 \geq 5 \) - \( n(1 - p) = 48 \times .22 = 10.56 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ### Middle Managers (n = 60) - \( np = 60 \times .78 = 46.8 \geq 5 \) - \( n(1 - p) = 60 \times .22 = 13.2 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ### Junior Associates (n = 55) - \( np = 55 \times .78 = 42.9 \geq 5 \) - \( n(1 - p) = 55 \times .22 = 12.1 \geq 5 \) - **Result:** Conditions satisfied → **Normal approximation valid** ## **Conclusion** Since all six samples satisfy both conditions (\( np \geq 5 \) and \( n(1 - p) \geq 5 \)), the binomial distribution can be reliably approximated by the normal distribution for each group's sample proportion.

# Validation of Normal Approximation for Multiple Samples ## Introduction For each sample, we verify whether the normal approximation to the binomial distribution is appropriate based on the following conditions: - \( np \geq 5 \) - \( n(1 - p) \geq 5 \) ## Sample Analyses ### Primary Sample (n = 65) - \( np = 65 \times 0.78 = 50.7 \geq 5 \) - \( n(1 - p) = 65 \times 0.22 = 14.3 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ### Secondary Sample (n = 52) - \( np = 52 \times 0.78 = 40.56 \geq 5 \) - \( n(1 - p) = 52 \times 0.22 = 11.44 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ### Auxiliary Sample (n = 71) - \( np = 71 \times 0.78 = 55.38 \geq 5 \) - \( n(1 - p) = 71 \times 0.22 = 15.62 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ### Ad Hoc Group (n = 48) - \( np = 48 \times 0.78 = 37.44 \geq 5 \) - \( n(1 - p) = 48 \times 0.22 = 10.56 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ### Middle Managers (n = 60) - \( np = 60 \times 0.78 = 46.8 \geq 5 \) - \( n(1 - p) = 60 \times 0.22 = 13.2 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ### Junior Associates (n = 55) - \( np = 55 \times 0.78 = 42.9 \geq 5 \) - \( n(1 - p) = 55 \times 0.22 = 12.1 \geq 5 \) - Result: Conditions satisfied → Normal approximation valid ## Conclusion All six samples satisfy both conditions (\( np \geq 5 \) and \( n(1 - p) \geq 5 \)). Therefore, the binomial distribution can be reliably approximated by the normal distribution for each group's sample proportion.