6.3 Measures of centre and variability

Centre describes a typical value, while variability describes how widely observations differ. Both are needed to characterize numerical data.

6.3.1 Mean, median, and mode

Sample mean. The arithmetic average \(\bar x=(\sum_{i=1}^n x_i)/n\).

The median is the middle ordered value, or the average of the two middle values when \(n\) is even. The mode is the most frequent value. Mean uses every observation and is sensitive to extremes; median depends mainly on order and is resistant.

Worked Example: Calculating centre

For turbidity values \(0.20,0.24,0.25,0.31,0.80\), the mean is \(1.80/5=0.36\) NTU and the median is 0.25 NTU. The high value pulls the mean upward.

Try It. Find mean and median of \(2,3,3,4,8\).
Check Your Work Mean 4; median 3.

Worked Example: Updating a mean

Five readings have mean 12, so their sum is 60. Adding a sixth reading of 18 gives new mean \((60+18)/6=13\).

Try It. Four values have mean 9. Add a fifth value of 14 and find the new mean.
Check Your Work Original sum 36; new mean \(50/5=10\).

6.3.2 Variance and standard deviation

Sample standard deviation. The square root of sample variance, \[s=\sqrt{\frac{\sum_{i=1}^n(x_i-\bar x)^2}{n-1}}.\]

Variance uses squared deviations and has squared units. Standard deviation returns to the original measurement units and describes typical distance from the mean. Population variance uses divisor \(N\) instead of \(n-1\).

Worked Example: Computing sample variability

For \(2,4,6\), the mean is 4. Squared deviations are 4, 0, and 4. Thus \(s^2=8/(3-1)=4\) and \(s=2\).

Try It. Find the sample standard deviation of \(5,5,5\).
Check Your Work All deviations are zero, so \(s=0\).

Worked Example: Comparing consistency

Two sensors both average 1.0 mg/L. Sensor A has \(s=0.03\) mg/L and Sensor B has \(s=0.18\) mg/L. Sensor A’s readings are more tightly clustered around their mean.

Try It. Which is more consistent: \(s=2.1\) kPa or \(s=0.8\) kPa?
Check Your Work The readings with \(s=0.8\) kPa.
Applied Problem: Comparing two turbidity periods. Period A readings are \(0.20,0.25,0.30\) NTU. Period B readings are \(0.10,0.25,0.40\) NTU. Calculate each mean and sample standard deviation, then compare centre and consistency.
Check Your Work Both means are 0.25 NTU. For A, \(s=0.05\) NTU. For B, \(s=0.15\) NTU. The periods have equal centres, but A is more consistent.

6.3.3 Practice Problems

  1. Find the mean of \(4,6,8\).
    Check Your Work 6.
  2. Find the median of \(2,5,7,11\).
    Check Your Work 6.
  3. Find the mode of \(1,2,2,4\).
    Check Your Work 2.
  4. Find sample variance and standard deviation of \(1,3,5\).
    Check Your Work \(s^2=4\), \(s=2\).
  5. Which has the same units as the data, variance or standard deviation?
    Check Your Work Standard deviation.
  6. Find the mean flow of 18, 20, 22, and 24 L/s.
    Check Your Work 21 L/s.
  7. Chlorine readings are 0.6, 0.6, 0.7, 0.8, 2.1. Which centre is more resistant?
    Check Your Work The median, 0.7 mg/L.
  8. Two pressure datasets have equal mean, with standard deviations 4 kPa and 11 kPa. Which is more variable?
    Check Your Work The dataset with 11 kPa.