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Bus ticket counter: mean, SD, CV, CI

Short answer

Sample mean = 45.70 passengers, sample standard deviation = 12.75, coefficient of variation = 27.90%, standard error = 1.65. The 95% confidence limits are (42.47, 48.93) and 99% limits (41.46, 49.94) passengers per hour.

Following data gives the sample records of number of passenger take ticket at the counter of Bus during one hour period. 22 58 32 36 62 57 25 45 23 37 64 56 46 60 29 49 63 36 26 58 60 26 58 58 29 43 53 36 45 22 52 43 45 31 45 39 35 38 30 60 58 42 54 62 52 42 65 58 51 60 53 45 31 53 22 53 51 52 47 59 Find the i. Sample mean of number of passenger ii. Sample standard deviation and coefficient of variation. iii. Standard error of the sample mean. iv. Find the 95% and 99% confidence limit of sample mean.

The data are ungrouped, n = 60.

i. Sample mean

Σx = 2742
x̄ = Σx/n = 2742/60 = 45.700

x̄ ≈ 45.70 passengers per hour.

ii. Standard deviation and coefficient of variation

Σx² = 134904
Σ(x − x̄)² = Σx² − n·x̄² = 134904 − 60×45.700² = 9594.600
s² = Σ(x − x̄)²/(n−1) = 9594.600/59 = 162.620
s  = 12.752
CV = (s/x̄) × 100 = (12.752/45.700) × 100 = 27.90%

s ≈ 12.75, CV ≈ 27.90%.

iii. Standard error of the mean

SE = s/√n = 12.752/√60 = 1.646

SE ≈ 1.65.

iv. Confidence limits

Since n = 60 (large), the normal critical values are used: x̄ ± z·SE.

95%: z = 1.960, margin = 1.960 × 1.646 = 3.227
     limits = 45.700 ± 3.227 = (42.47, 48.93)
99%: z = 2.576, margin = 2.576 × 1.646 = 4.241
     limits = 45.700 ± 4.241 = (41.46, 49.94)

Answer: 95% limits: (42.47, 48.93); 99% limits: (41.46, 49.94) passengers per hour.

✓ Answer checked by Claude Sonnet 5.5

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