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1. Primary Sample (n = 65) Check Normality Conditions: ? ? = 65 × 0.78 np=65×0.78   = 50.7   ≥ 5 ? ( 1 − ? ) = 65 × ( 1 − 0.78 ) n(1−p)=65×(1−0.78)   = 65 × 0.22   = 14.3   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 2. Secondary Sample (n = 52) Check Normality Conditions: ? ? = 52 × 0.78 np=52×0.78   = 40.56   ≥ 5 ? ( 1 − ? ) = 52 × ( 1 − 0.78 ) n(1−p)=52×(1−0.78)   = 52 × 0.22   = 11.44   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 3. Auxiliary Sample (n = 71) Check Normality Conditions: ? ? = 71 × 0.78 np=71×0.78   = 55.38   ≥ 5 ? ( 1 − ? ) = 71 × ( 1 − 0.78 ) n(1−p)=71×(1−0.78)   = 71 × 0.22   = 15.62   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 4. Ad Hoc Group (n = 48) Check Normality Conditions: ? ? = 48 × 0.78 np=48×0.78   = 37.44   ≥ 5 ? ( 1 − ? ) = 48 × ( 1 − 0.78 ) n(1−p)=48×(1−0.78)   = 48 × 0.22   = 10.56   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 5. Middle Managers (n = 60) Check Normality Conditions: ? ? = 60 × 0.78 np=60×0.78   = 46.8   ≥ 5 ? ( 1 − ? ) = 60 × ( 1 − 0.78 ) n(1−p)=60×(1−0.78)   = 60 × 0.22   = 13.2   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 6. Junior Associates (n = 55) Check Normality Conditions: ? ? = 55 × 0.78 np=55×0.78   = 42.9   ≥ 5 ? ( 1 − ? ) = 55 × ( 1 − 0.78 ) n(1−p)=55×(1−0.78)   = 55 × 0.22   = 12.1   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. ✅ Final Conclusion: For all six samples, ? ? ≥ 5 and ? ( 1 − ? ) ≥ 5 np≥5andn(1−p)≥5 Hence, the binomial distribution can be approximated by the normal distribution for each sample.

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1. Primary Sample (n = 65) Check Normality Conditions: ? ? = 65 × 0.78 np=65×0.78   = 50.7   ≥ 5 ? ( 1 − ? ) = 65 × ( 1 − 0.78 ) n(1−p)=65×(1−0.78)   = 65 × 0.22   = 14.3   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 2. Secondary Sample (n = 52) Check Normality Conditions: ? ? = 52 × 0.78 np=52×0.78   = 40.56   ≥ 5 ? ( 1 − ? ) = 52 × ( 1 − 0.78 ) n(1−p)=52×(1−0.78)   = 52 × 0.22   = 11.44   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 3. Auxiliary Sample (n = 71) Check Normality Conditions: ? ? = 71 × 0.78 np=71×0.78   = 55.38   ≥ 5 ? ( 1 − ? ) = 71 × ( 1 − 0.78 ) n(1−p)=71×(1−0.78)   = 71 × 0.22   = 15.62   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 4. Ad Hoc Group (n = 48) Check Normality Conditions: ? ? = 48 × 0.78 np=48×0.78   = 37.44   ≥ 5 ? ( 1 − ? ) = 48 × ( 1 − 0.78 ) n(1−p)=48×(1−0.78)   = 48 × 0.22   = 10.56   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 5. Middle Managers (n = 60) Check Normality Conditions: ? ? = 60 × 0.78 np=60×0.78   = 46.8   ≥ 5 ? ( 1 − ? ) = 60 × ( 1 − 0.78 ) n(1−p)=60×(1−0.78)   = 60 × 0.22   = 13.2   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. 6. Junior Associates (n = 55) Check Normality Conditions: ? ? = 55 × 0.78 np=55×0.78   = 42.9   ≥ 5 ? ( 1 − ? ) = 55 × ( 1 − 0.78 ) n(1−p)=55×(1−0.78)   = 55 × 0.22   = 12.1   ≥ 5 ✅ Both conditions are satisfied → normal approximation is valid. ✅ Final Conclusion: For all six samples, ? ? ≥ 5 and ? ( 1 − ? ) ≥ 5 np≥5andn(1−p)≥5 Hence, the binomial distribution can be approximated by the normal distribution for each sample.

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Asked by: nayan
Created at: 2025-10-27 06:13:02

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# Normality for Multiple SamplesThis document verifies whether the normal approximation the binomial distribution is valid for six different samples based on normality conditions. --- ## Normality Check Criteria For each sample,: - \( np \geq5 \) - \( n(1 - p) \geq 5 \) where: - \( n \) = sample size - \( p = .78 \) (proportion) --- ## Sample-by-Sample Analysis ### 1. Primary Sample (n = 65) - \( np = 65 \times .78 = 50.7 \geq 5 \) - \( n(1 - p) = 65 \times .22 = 14.3 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ### 2. Secondary Sample (n = 52) - \( np = 52 \times .78 = 40.56 \geq 5 \) - \( n(1 - p) = 52 \times .22 = 11.44 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ### 3. Auxiliary Sample (n = 71) - \( np = 71 \times .78 = 55.38 \geq 5 \) - \( n(1 - p) = 71 \times .22 = 15.62 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ### 4. Ad Hoc Group (n = 48) - \( np = 48 \times .78 = 37.44 \geq 5 \) - \( n(1 - p) = 48 \times .22 = 10.56 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ### 5. Middle Managers (n = 60) - \( np = 60 \times .78 = 46.8 \geq 5 \) - \( n(1 - p) = 60 \times .22 = 13.2 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ### 6. Junior Associates (n = 55) - \( np = 55 \times .78 = 42.9 \geq 5 \) - \( n(1 - p) = 55 \times .22 = 12.1 \geq 5 \) **Result:** Both conditions satisfied → Normal approximation valid. --- ## **Final Conclusion** For **all six samples**, the conditions: \[ np \geq 5 \quad \text{and} \quad n(1 - p) \geq 5 \] are satisfied. Therefore, the binomial distribution can be **approximated by the normal distribution** for each sample.

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