Question

Rainwater was collected in water collectors at thirty different sites near an industrial basin and the amount of acidity (pH level) was measured. The mean and standard deviation of the values are 4.9 and 1.5 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 0.2 pH units to all of the values and then multiply the result by 1.3. Find the mean and standard deviation of the corrected pH measurements.

Answer 1

Answer:

The mean and standard deviation of the corrected pH measurements are 6.63 and 3.8025 respectively.

Step-by-step explanation:

We can correct the values of the mean and standard deviation using the properties of the mean and the variance.

To the original value X we have to add 0.2 and multiply then by 1.3 to calculate the new and corrected value Y:

$Y=1.3(X+0.2)$

The mean and standard deviation of the original value X are 4.9 and 1.5 respectively.

Then, we can apply the properties of the mean as:

$E(Y)=E(1.3(X+0.2))=1.3E(X+0.2)=1.3E(X)+1.3*0.2\\\\E(Y)=1.3E(X)+0.26\\\\E(Y)=1.3*4.9+0.26=6.37+0.26=6.63$

For the standard deviation, we apply the properties of variance:

$V(Y)=V(1.3(X+0.2))\\\\V(Y)=1.3^2\cdot V(X+0.2)\\\\V(Y)=1.69\cdot V(X)\\\\V(Y)=1.69\cdot 1.5^2=1.69\cdot 2.25=3.8025$

The properties that have been applied are:

$1.\,E(aX)=aE(X)\\\\ 2.\,E(X+b)=E(X)+b\\\\3.\,V(aX)=a^2V(X)\\\\4.\,V(X+b)=V(X)+0$