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Mean, variance, CV of preschool ages

Short answer

Sample mean x̄ = 39.08 months, variance s² = 35.83 months², standard deviation s ≈ 5.99 months. The coefficient of variation CV = (s/x̄)×100 ≈ 15.3% measures relative variability.

The following data are the ages (in months) at which n = 50 children were first enrolled in a preschool. 38 40 30 35 39 47 35 34 43 41 32 34 41 30 46 55 39 33 32 32 42 50 37 39 33 40 48 36 31 36 36 41 43 48 40 35 40 30 46 37 45 42 41 36 50 45 38 46 36 31 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 = Sxx/(n − 1) with Sxx = Σx² − (Σx)²/n, and the coefficient of variation is CV = (s / x̄) × 100%.

a) Mean, variance and standard deviation

n = 50. From the data, Σx = 1954 and Σx² = 78,118.

x̄  = 1954 / 50 = 39.08 months
Sxx = 78,118 − (1954)²/50
    = 78,118 − 76,362.32
    = 1755.68
s²  = 1755.68 / 49 = 35.83 months²
s   = √35.83 = 5.99 months

b) Relative variability

The coefficient of variation measures the spread relative to the mean:

CV = (5.986 / 39.08) × 100 = 15.32 %

Answer: x̄ = 39.08 months, s² ≈ 35.83 months², s ≈ 5.99 months, CV ≈ 15.3 %. The spread is about 15% of the mean age, which is moderate.

✓ Answer checked by Claude Sonnet 5.5

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