Rainwater was collected in water collectors at 30 different sites near an industrial complex, and the amount of acidity (pH level) was measured. The mean and standard deviation of the values are 4.60 and 1.10, respectively. When the pH meter was recalibrated back at the laboratory, it was found to be in error. The error can be corrected by adding 0.1 pH units to all of the values and then multiplying the result by 1.2. The mean and standard deviation of the corrected pH measurements are (a) 5.64, 1.44 (b) 5.64, 1.32 (c) 5.40, 1.44 (d) 5.40, 1.32 (e) 5.64, 1.20
The Correct Answer and Explanation is:
Given Data:
- Number of sites: 30 (not directly relevant for this calculation)
- Original mean xˉ=4.60\bar{x} = 4.60
- Original standard deviation s=1.10s = 1.10
- Correction formula for each pH value xx:
xcorrected=(x+0.1)×1.2x_{corrected} = (x + 0.1) \times 1.2
Goal:
Find the corrected mean and corrected standard deviation after applying the correction to all measurements.
Step 1: Corrected Mean
The mean of a transformed variable Y=aX+bY = aX + b (where aa and bb are constants) is: Yˉ=aXˉ+b\bar{Y} = a \bar{X} + b
In this case, the transformation is: xcorrected=1.2×(x+0.1)=1.2x+1.2×0.1=1.2x+0.12x_{corrected} = 1.2 \times (x + 0.1) = 1.2x + 1.2 \times 0.1 = 1.2x + 0.12
So, a=1.2,b=0.12a = 1.2, \quad b = 0.12
The corrected mean is: xˉcorrected=1.2×4.60+0.12=5.52+0.12=5.64\bar{x}_{corrected} = 1.2 \times 4.60 + 0.12 = 5.52 + 0.12 = 5.64
Step 2: Corrected Standard Deviation
The standard deviation after a linear transformation Y=aX+bY = aX + b changes as: sY=∣a∣sXs_Y = |a| s_X
Note that adding a constant bb does not affect the standard deviation.
Here, scorrected=∣1.2∣×1.10=1.32s_{corrected} = |1.2| \times 1.10 = 1.32
Final answer:
- Corrected mean =5.64= 5.64
- Corrected standard deviation =1.32= 1.32
This corresponds to option (b) 5.64, 1.32.
Explanation:
When all data points are adjusted by first adding 0.1 to each and then multiplying the result by 1.2, the transformation can be considered linear: Y=aX+bY = aX + b with a=1.2a = 1.2 and b=0.12b = 0.12.
- The mean shifts accordingly: scaling the original mean by 1.2 and then adding 0.12.
- The standard deviation is scaled by the absolute value of 1.2, since adding a constant does not affect variability.
Thus, the variability in the measurements increases proportionally to the multiplication factor 1.2, and the mean shifts based on both the addition and multiplication. This results in a corrected mean of 5.64 and a corrected standard deviation of 1.32.
