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Sample mean, variance, CV: patients' sleep hours

Short answer

Sample mean = 6.71 hours, variance = 16.21 hours², standard deviation = 4.03 hours. The coefficient of variation measuring relative variability is (4.03/6.71)×100% = 60%.

The following table shows the number of hours 45 hospital patients slept following the administration of a certain anesthetic. 7 10 12 4 8 7 3 8 5 12 11 3 8 1 1 13 10 4 4 5 5 8 7 7 3 2 3 8 13 1 7 17 3 4 5 5 3 1 17 10 4 7 7 11 8 a) Find sample mean, sample variance and sample standard deviation. b) Compute a value that measures the amount of variability relative to the value of mean.

The sample mean is x̄ = Σx / n, the sample variance is s² = Σ(x − x̄)² / (n − 1), the sample standard deviation is s = √s², and the coefficient of variation is CV = (s / x̄) × 100%.

a) Mean, variance and standard deviation

Data: n = 45 values.

  • Σx = 302
  • Σx² = 2740

Mean:

x̄ = Σx / n = 302 / 45 = 6.711 hours

Sum of squares about the mean (corrected sum of squares):

Sxx = Σx² − (Σx)²/n
    = 2740 − (302)²/45
    = 2740 − 2026.756
    = 713.244

Variance and standard deviation:

s² = Sxx / (n − 1) = 713.244 / 44 = 16.210 hours²
s  = √16.210 = 4.026 hours

b) Relative variability

The measure that compares spread with the mean is the coefficient of variation:

CV = (s / x̄) × 100 = (4.026 / 6.711) × 100 = 59.99 %

Answer: x̄ ≈ 6.71 hours, s² ≈ 16.21 hours², s ≈ 4.03 hours, CV ≈ 60 %. The standard deviation is about 60% of the mean, so the sleeping hours vary a lot between patients.

✓ Answer checked by Claude Sonnet 5.5

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